Chapter 7: Integrals

We use the Inspection Method to find an antiderivative of $\sin 2 x$.

We know that $\frac{d}{d x}(\cos 2 x)=-2 \sin 2 x$

Dividing by $-2, \frac{-1}{2} \frac{d}{d x}(\cos 2 x)=\sin 2 x$

or $\frac{d}{d x}\left(\frac{-1}{2} \cos 2 x\right)=\sin 2 x$

∴ By definition, an antiderivative of $\sin 2 x$ is $\frac{-1}{2} \cos 2 x$.

Note: The complete antiderivative or integral of $\sin 2 x$ is $\frac{-1}{2} \cos 2 x+c$. For different values of $c$, we get different antiderivatives. We drop $c$ when identifying just one antiderivative.

We use the Inspection Method to find an antiderivative of $\cos 3 x$.

We know that $\frac{d}{d x}(\sin 3 x)=3 \cos 3 x$

Dividing by $3, \frac{1}{3} \frac{d}{d x}(\sin 3 x)=\cos 3 x$ or $\frac{d}{d x}\left(\frac{1}{3} \sin 3 x\right)=\cos 3 x$

∴ By definition, an antiderivative of $\cos 3 x$ is $\frac{1}{3} \sin 3 x$.

(Refer to the note after Q.1 regarding omission of $c$.)

We apply the Inspection Method to find an antiderivative of $e^{2 x}$.

We know that $\frac{d}{d x} e^{2 x}=e^{2 x} \frac{d}{d x}(2 x)=2 e^{2 x}$

Dividing by $2, \frac{1}{2} \frac{d}{d x} e^{2 x}=e^{2 x} \quad$ or $\quad \frac{d}{d x}\left(\frac{1}{2} e^{2 x}\right)=e^{2 x}$

Therefore, an antiderivative of $e^{2 x}$ is $\frac{1}{2} e^{2 x}$.

Our aim is to find an antiderivative of $(a x+b)^2$.

We know that $\frac{d}{d x}(a x+b)^3=3(a x+b)^2 \frac{d}{d x}(a x+b)=3(a x+b)^2 a$.

Dividing by $3 a, \frac{1}{3 a} \frac{d}{d x}(a x+b)^3=(a x+b)^2$

or

$\therefore \quad$ An antiderivative of $(a x+b)^2$ is $\frac{1}{3 a}(a x+b)^3$.

We use the Inspection Method to find an antiderivative of $\sin 2 x-4 e^{3 x}$.

We know that $\quad \frac{d}{d x}(\cos 2 x)=-2 \sin 2 x$

Dividing by $-2, \frac{d}{d x}\left(\frac{-1}{2} \cos 2 x\right)=\sin 2 x$

Again $\frac{d}{d x} e^{3 x}=3 e^{3 x}$

Multiplying by $-4, \frac{d}{d x}\left(\frac{-4}{3} e^{3 x}\right)=-4 e^{3 x}$

Adding equations (i) and (ii)

or

Therefore, an antiderivative of $\sin 2 x-4 e^{3 x}$ is $\frac{-1}{2} \cos 2 x-\frac{4}{3} e^{3 x}$.

Evaluate the following integrals in Exercises 6 to 11.

$\int\left(4 e^{3 x}+1\right) d x=\int 4 e^{3 x} d x+\int 1 d x$

$\int x^2\left(1-\frac{1}{x^2}\right) d x=\int\left(x^2-\frac{x^2}{x^2}\right) d x=\int\left(x^2-1\right) d x$

$\int\left(a x^2+b x+c\right) d x=\int a x^2 d x+\int b x d x+\int c d x$

Here $c_1$ is the arbitrary constant of integration.

$\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 d x$

$\int \frac{x^3+5 x^2-4}{x^2} d x=\int\left(\frac{x^3}{x^2}+\frac{5 x^2}{x^2}-\frac{4}{x^2}\right) d x$

Evaluate the following integrals in Exercises 12 to 16.

$\int \frac{x^3+3 x+4}{\sqrt{x}} d x=\int\left(\frac{x^3}{x^{1 / 2}}+\frac{3 x}{x^{1 / 2}}+\frac{4}{x^{1 / 2}}\right) d x$

$\quad \int \frac{x^3-x^2+x-1}{x-1} d x=\int \frac{x^2(x-1)+(x-1)}{x-1} d x$

Evaluate the following integrals in Exercises 17 to 20.

$\int\left(2 x^2-3 \sin x+5 \sqrt{x}\right) d x$

$\int \sec x(\sec x+\tan x) d x=\int\left(\sec ^2 x+\sec x \tan x\right) d x$

$\int \frac{\sec ^2 x}{\operatorname{cosec}^2 x} d x=\int \frac{\frac{1}{\cos ^2 x}}{\frac{1}{\sin ^2 x}} d x=\int \frac{\sin ^2 x}{\cos ^2 x} d x$

Note: In a similar way, $\int \cot ^2 x d x=\int\left(\operatorname{cosec}^2 x-1\right) d x$

$\int \frac{2-3 \sin x}{\cos ^2 x} d x=\int\left(\frac{2}{\cos ^2 x}-\frac{3 \sin x}{\cos ^2 x}\right) d x$

The antiderivative of the $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$

Hence, Option (C) is the correct answer.

Given: $\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4}$ and $f(2)=0$

∴ By definition of antiderivative (i.e., Integral),

To find $\boldsymbol{c}$. Let us make use of $f(2)=0$ (given)

Putting $x=2$ on both sides of (i),

( $\because f(2)=0$ (given))

or $c+\frac{129}{8}=0 \quad$ or $c=\frac{-129}{8}$

Substituting $c=\frac{-129}{8}$ back into (i), $f(x)=x^4+\frac{1}{\left(x^3\right)}-\frac{129}{8}$

Hence, Option (A) is the correct answer.

We now evaluate $\int \frac{2 x}{1+x^2} d x$

Put $1+x^2=t$. Therefore $2 x=\frac{d t}{d x}$ or $2 x d x=d t$

Putting $t=1+x^2, \quad=\log \left|1+x^2\right|+c=\log \left(1+x^2\right)+c$.

We now evaluate $\int \frac{(\log x)^2}{x} d x$

Put $\log x=t$. Therefore $\frac{1}{x}=\frac{d t}{d x} \quad \Rightarrow \quad \frac{d x}{x}=d t$

Putting $t=\log x,=\frac{1}{3}(\log x)^3+c$.

We now evaluate $\int \frac{1}{x+x \log x} d x=\int \frac{1}{x(1+\log x)} d x$

Put $1+\log x=t . \quad$ Therefore $\frac{1}{x}=\frac{d t}{d x} \Rightarrow \frac{d x}{x}=d t$

Putting $t=1+\log x, \log |1+\log x|+c$.

We now evaluate $\int \sin x \sin (\cos x) d x=-\int \sin (\cos x)(-\sin x) d x$

Put $\cos x=t$. Therefore $-\sin x=\frac{d t}{d x}$

Putting $t=\cos x,=\cos (\cos x)+c$.

We now evaluate $\int \sin (a x+b) \cos (a x+b) d x$

We now evaluate $\int \sqrt{a x+b} d x=\int(a x+b)^{1 / 2} d x$

We now evaluate $\int x \sqrt{x+2} d x$

OR

We now evaluate $\int x \sqrt{x+2} d x$

Put $\sqrt{\text { Linear }}=t$, i.e., $\sqrt{x+2}=t$.

Squaring $x+2=t^2 \quad\left(\Rightarrow x=t^2-2\right)$

Putting $t=\sqrt{x+2}, \quad=\frac{2}{5}(\sqrt{x+2})^5-\frac{4}{3}(\sqrt{x+2})^3+c$

$\left.=\frac{2}{5}(x+2)^{1 / 2}\right)^5-\frac{4}{3}\left((x+2)^{1 / 2}\right)^3+c=\frac{2}{5}(x+2)^{5 / 2}-\frac{4}{3}(x+2)^{3 / 2}+c$.

We now evaluate $\int x \sqrt{1+2 x^2} d x$

Let $\mathrm{I}=\int x \sqrt{1+2 x^2} d x=\frac{1}{4} \int \sqrt{1+2 x^2}(4 x d x)$

Put $1+2 x^2=t$. Therefore $4 x=\frac{d t}{d x} \quad$ or $\quad 4 x d x=d t$

Therefore, from (i), I $=\frac{1}{4} \int \sqrt{t} d t=\frac{1}{4} \int t^{1 / 2} d t$

Putting $t=1+2 x^2,=\frac{1}{6}\left(1+2 x^2\right)^{3 / 2}+c$.

Integrate the functions in Exercises 9 to 17:

Let $\mathrm{I}=\int(4 x+2) \sqrt{x^2+x+1} d x=\int 2(2 x+1) \sqrt{x^2+x+1} d x$

Put $x^2+x+1=t$. Therefore $(2 x+1)=\frac{d t}{d x}$

Putting $t=x^2+x+1, \mathrm{I}=\frac{4}{3}\left(x^2+x+1\right)^{3 / 2}+c$.

Let $\mathrm{I}=\int \frac{1}{x-\sqrt{x}} d x$

Put $\sqrt{\text { Linear }}=t$, i.e., $\sqrt{x}=t$

Squaring: $x=t^2$. Therefore $\frac{d x}{d t}=2 t \quad$ or $\quad d x=2 t d t$

Putting $t=\sqrt{x}, \mathrm{I}=2 \log |\sqrt{x}-1|+c$.

Let I $=\int \frac{x}{\sqrt{x+4}} d x$

OR

Put $\sqrt{\text { Linear }}=\boldsymbol{t}$, i.e., $\sqrt{x+4}=t$.

Squaring $x+4=t^2 \Rightarrow x=t^2-4$.

Therefore $\frac{d x}{d t}=2 t \quad$ or $\quad d x=2 t d t$

Putting $t=\sqrt{x+4},=\frac{2}{3} \sqrt{x+4}(x+4-12)+c$

Let $\mathrm{I}=\int\left(x^3-1\right)^{1 / 3} x^5 d x=\int\left(x^3-1\right)^{1 / 3} x^3 x^2 d x$

Put $x^3-1=t \quad \Rightarrow \quad x^3=t+1$

∴ From $(i), \quad \mathrm{I}=\frac{1}{3} \int t^{1 / 3}(t+1) d t$

Putting $t=x^3-1,=\frac{1}{7}\left(x^3-1\right)^{7 / 3}+\frac{1}{4}\left(x^3-1\right)^{4 / 3}+c$.

Let $\mathrm{I}=\int \frac{x^2}{\left(2+3 x^3\right)^3} d x$

Put $2+3 x^3=t$. Therefore $9 x^2=\frac{d t}{d x} \Rightarrow 9 x^2 d x=d t$

Therefore, from (i), I $=\frac{1}{9} \int t^{-3} d t \quad=\frac{1}{9} \frac{t^{-2}}{-2}+c=\frac{-1}{18 t^2}+c$

Putting $t=2+3 x^3 ;=\frac{-1}{18\left(2+3 x^3\right)^2}+c$.

Let $\mathrm{I}=\int \frac{1}{x(\log x)^m} d x(x>0) \Rightarrow \mathrm{I}=\int \frac{\frac{1}{x} d x}{(\log x)^m}$

Put $\log x=t$. Therefore $\frac{1}{x}=\frac{d t}{d x} \Rightarrow \frac{d x}{x}=d t$

Therefore, from (i), I $=\int \frac{d t}{t^m}=\int t^{-m} d t=\frac{t^{-m+1}}{-m+1}+c$

(Assuming $m \neq 1$ )

Putting $t=\log x,=\frac{(\log x)^{1-m}}{1-m}+c$.

Let $\mathrm{I}=\int \frac{x}{9-4 x^2} d x=\frac{-1}{8} \int \frac{-8 x}{9-4 x^2} d x$

Put $9-4 x^2=t$. Therefore $-8 x=\frac{d t}{d x} \Rightarrow-8 x d x=d t$

From equation (i), I = $\frac{-1}{8} \int \frac{d t}{t}=\frac{-1}{8} \int \frac{1}{t} d t \quad=\frac{-1}{8} \log |t|+c$

Putting $t=9-4 x^2,=\frac{-1}{8} \log \left|9-4 x^2\right|+c$.

$\int e^{2 x+3} d x=\frac{e^{2 x+3}}{2 \rightarrow \text { Coeff. of } x}+c$

Let $\mathrm{I}=\int \frac{x}{\left(e^{x^2}\right)} d x=\frac{1}{2} \int \frac{2 x}{\left(e^{x^2}\right)} d x$

Put $x^2=t$. Therefore $2 x=\frac{d t}{d x} \Rightarrow 2 x d x=d t$.

Putting $t=x^2, \mathrm{I}=\frac{-1}{2\left(e^{x^2}\right)}+c$.

Integrate the functions in Exercises 18 to 26:

Let $\mathrm{I}=\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x$

Put $\tan ^{-1} x=t$.

Therefore, from (i), $\mathrm{I}=\int e^t d t=e^t+c=e^{\tan ^{-1} x}+c$.

Let $\mathrm{I}=\int \frac{e^{2 x}-1}{e^{2 x}+1} d x$

Multiplying every term in integrand by $e^{-x}$,

Put denominator $e^x+e^{-x}=t$

$\therefore \quad e^x+e^{-x} \frac{d}{d x}(-x)=\frac{d t}{d x} \quad \Rightarrow \quad\left(e^x-e^{-x}\right) d x=d t$

Therefore, from (i), I $=\int \frac{d t}{t}=\int \frac{1}{t} d t=\log |t|+c$

Putting $t=e^x+e^{-x}, \mathrm{I}=\log \left|e^x+e^{-x}\right|+c$ or $\mathrm{I}=\log \left(e^x+e^{-x}\right)+c$

Let $\mathrm{I}=\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x \quad=\frac{1}{2} \int \frac{2\left(e^{2 x}-e^{-2 x}\right)}{e^{2 x}+e^{-2 x}} d x$

Put denominator $e^{2 x}+e^{-2 x}=t$

Putting $t=e^{2 x}+e^{-2 x},=\frac{1}{2} \log \left|e^{2 x}+e^{-2 x}\right|+c=\frac{1}{2} \log \left(e^{2 x}+e^{-2 x}\right)+c$

$\int \tan ^2(2 x-3) d x=\int\left(\sec ^2(2 x-3)-1\right) d x\left(\because \tan ^2 \theta=\sec ^2 \theta-1\right)$

$\int \sec ^2(7-4 x) d x=\frac{\tan (7-4 x)}{-4 \rightarrow \text { Coeff. of } x}+c$

$=\frac{-1}{4} \tan (7-4 x)+c$.

Let $\mathrm{I}=\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$

Put $\sin ^{-1} x=t \quad \therefore \quad \frac{1}{\sqrt{1-x^2}}=\frac{d t}{d x} \quad \Rightarrow \frac{d x}{\sqrt{1-x^2}}=d t$

Therefore, from (i), I $=\int t d t=\frac{t^2}{2}+c$

Putting $t=\sin ^{-1} x, \mathrm{I}=\frac{1}{2}\left(\sin ^{-1} x\right)^2+c$.

Let $\mathrm{I}=\int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x=\int \frac{2 \cos x-3 \sin x}{2(2 \sin x+3 \cos x)} d x$

Put DENOMINATOR $2 \sin x+3 \cos x=t$

$\therefore \quad 2 \cos x-3 \sin x=\frac{d t}{d x} \Rightarrow(2 \cos x-3 \sin x) d x=d t$

Therefore, from (i), I $=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log |t|+c$.

Putting $t=2 \sin x+3 \cos x,=\frac{1}{2} \log |2 \sin x+3 \cos x|+c$.

Let $\mathrm{I}=\int \frac{1}{\cos ^2 x(1-\tan x)^2} d x=\int \frac{\sec ^2 x}{(1-\tan x)^2} d x$

Put $1-\tan x=t$.

Let $\mathrm{I}=\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$

Put $\sqrt{\text { Linear }}=t$, i.e., $\quad \sqrt{x}=t$

Squaring, $x=t^2$. Therefore $\frac{d x}{d t}=2 t \quad \therefore \quad d x=2 t d t$

Therefore, from (i), I $=\int \frac{\cos t}{t} 2 t d t=2 \int \cos t d t=2 \sin t+c$

Putting $t=\sqrt{x}, \mathrm{I}=2 \sin \sqrt{x}+c$.

Integrate the functions in Exercises 27 to 37:

Let $\mathrm{I}=\int \sqrt{\sin 2 x} \cos 2 x d x=\frac{1}{2} \int \sqrt{\sin 2 x}(2 \cos 2 x d x)$

Put $\boldsymbol{\operatorname { s i n }} \mathbf{2 x}=\boldsymbol{t}$

$\therefore \cos 2 x \frac{d}{d x}(2 x)=\frac{d t}{d x} \quad \Rightarrow 2 \cos 2 x d x=d t$

Let $\mathrm{I}=\int \frac{\cos x}{\sqrt{1+\sin x}} d x$

Put $1+\sin x=t$

$\therefore \quad \cos x=\frac{d t}{d x} \quad$ or $\quad \cos x d x=d t$

Let $\mathrm{I}=\int \cot x \log \sin x d x$

Put $\log \sin x=t$

$\therefore \quad \frac{1}{\sin x} \frac{d}{d x}(\sin x)=\frac{d t}{d x} \quad$ or $\quad \frac{1}{\sin x} \cos x=\frac{d t}{d x}$

or $\cot x d x=d t$

∴ From $(i), \mathrm{I}=\int t d t=\frac{t^2}{2}+c \quad=\frac{1}{2}(\log \sin x)^2+c$.

Let $\mathrm{I}=\int \frac{\sin x}{1+\cos x} d x \quad=-\int \frac{-\sin x}{1+\cos x} d x$

Put $1+\cos x=t$. Therefore $-\sin x=\frac{d t}{d x}$

$\therefore \quad-\sin x d x=d t$

∴ From $(i), \mathrm{I}=-\int \frac{d t}{t} \quad=-\log |t|+c$

Putting $t=1+\cos x,=-\log |1+\cos x|+c$.

Let $\mathrm{I}=\int \frac{\sin x}{(1+\cos x)^2} d x=-\int \frac{-\sin x d x}{(1+\cos x)^2}$

Put $1+\cos x=t$. Therefore $-\sin x=\frac{d t}{d x}$

Let $\mathrm{I}=\int \frac{1}{1+\cot x} d x=\int \frac{1}{1+\frac{\cos x}{\sin x}} d x=\int \frac{1}{\left(\frac{\sin x+\cos x}{\sin x}\right)} d x$

Adding and subtracting $\cos x$ in the numerator of integrand,

where $\mathrm{I}_1=\int \frac{\cos x-\sin x}{\sin x+\cos x} d x$

Put DENOMINATOR $\sin x+\cos x=t$

Note: Alternative solution for finding $\mathbf{I}_1$

Substituting the value of $\mathrm{I}_1$ into ( $i$ ), required integral

Let $\mathrm{I}=\int \frac{1}{1-\tan x} d x=\int \frac{1}{1-\frac{\sin x}{\cos x}} d x=\int \frac{1}{\left(\frac{\cos x-\sin x}{\cos x}\right)} d x$

Subtracting and adding $\sin x$ in the Numerator,

Note: Alternative solution for evaluating $\int \frac{-\sin x-\cos x}{\cos x-\sin x} d x$, put denominator $\cos x-\sin x=t$.

Let $\mathrm{I}=\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x=\int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x} \cos x \cos x} d x$

Put $\tan \boldsymbol{x}=\boldsymbol{t}$.

Let $\mathrm{I}=\int \frac{(1+\log x)^2}{x} d x$

Put $1+\log x=t$

Let $\mathrm{I}=\int \frac{(x+1)(x+\log x)^2}{x} d x$

Put $x+\log x=t$

$\therefore \quad 1+\frac{1}{x}=\frac{d t}{d x} \Rightarrow \frac{x+1}{x}=\frac{d t}{d x} \Rightarrow\left(\frac{x+1}{x}\right) d x=d t$

Therefore, from (i), $\mathrm{I}=\int t^2 d t=\frac{t^3}{3}+c$

Putting $t=x+\log x, \frac{1}{3}(x+\log x)^3+c$.

Let $\mathrm{I}=\int \frac{x^3 \sin \left(\tan ^{-1} x^4\right)}{1+x^8} d x=\frac{1}{4} \int \sin \left(\tan ^{-1} x^4\right) \cdot \frac{4 x^3}{1+x^8} d x$

Put $\left(\tan ^{-1} x^4\right)=t$

[Rule for $\int \sin (f(x)) f^{\prime}(x) d x$; put $f(x)=t$ ]

$\therefore \quad \frac{1}{1+\left(x^4\right)^2} \quad \frac{d}{d x} x^4=\frac{d t}{d x}\left[\because \quad \frac{d}{d x} \tan ^{-1} f(x)=\frac{1}{1+(f(x))^2} \frac{d}{d x} f(x)\right]$

$\Rightarrow \frac{4 x^3}{1+x^8} d x=d t$

Therefore, from (i),

For Exercises 38 and 39 — choose the correct option:

Let $\mathrm{I}=\int \frac{10 x^9+10^x \log _e 10}{x^{10}+10^x} d x$

Put $\boldsymbol{x}^{\mathbf{1 0}} \boldsymbol{+} \mathbf{1 0}^{\boldsymbol{x}} \boldsymbol{=} \boldsymbol{t}$

$\therefore\left(10 x^9+10^x \log _e 10\right) d x=d t \quad\left[\because \frac{d}{d x}\left(a^x\right)=a^x \log _e a\right]$

Therefore, from (i), I $=\int \frac{d t}{t}=\log |t|+c$

Putting $t=x^{10}+10^x, \mathrm{I}=\log \left|x^{10}+10^x\right|+c$

or $\mathrm{I}=\log \left(10^x+x^{10}\right)+c$.

Hence, Option (D) is the correct answer.

OR

$\int \frac{10 x^9+10^x \log _e 10}{x^{10}+10^x} d x=\int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c$

Hence, Option (D) is the correct answer.

$\int \frac{d x}{\sin ^2 x \cos ^2 x}=\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x} d x \quad\left[\because \quad 1=\sin ^2 x+\cos ^2 x\right]$

$\int \sin ^2(2 x+5) d x=\int \frac{1}{2}(1-\cos 2(2 x+5)) d x$

$\int \sin 3 x \cos 4 x d x=\frac{1}{2} \int 2 \sin 3 x \cos 4 x d x$

$\cos 2 x \cos 4 x \cos 6 x=\frac{1}{2}(2 \cos 6 x \cos 4 x) \cos 2 x$

Note: We know that $\sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta$

$\therefore 4 \sin ^3 \theta=3 \sin \theta-\sin 3 \theta$

Dividing by $4, \sin ^3 \theta=\frac{3}{4} \sin \theta-\frac{1}{4} \sin 3 \theta$

In a similar manner, $\cos ^3 \theta=\frac{3}{4} \cos \theta+\frac{1}{4} \cos 3 \theta$

We now evaluate $\int \sin ^3(2 x+1) d x$

We know by Equation (i) of above note that $\sin ^3 \theta=\frac{3}{4} \sin \theta-\frac{1}{4} \sin 3 \theta$ Putting $\theta=2 x+1$, we have

OR

To integrate $\sin ^n x$ where $n$ is odd, put $\cos x=t$.

Put $\cos (2 x+1)=t$

$\therefore \quad-\sin (2 x+1) \frac{d}{d x}(2 x+1)=\frac{d t}{d x} \therefore-2 \sin (2 x+1) d x=d t$

$\therefore \quad$ From $(i)$, the given integral $=\frac{-1}{2} \int\left(1-t^2\right) d t$

$\int \sin ^3 x \cos ^3 x d x=\int(\sin x \cos x)^3 d x$

We now evaluate $\int \sin ^3 x \cos ^3 x d x$, Put either $\sin x=t$ or $\cos x=t$. (The form of answer given in N.C.E.R.T. book II can be obtained by putting $\cos x=t$ )

$\sin x \sin 2 x \sin 3 x=\frac{1}{2}(2 \sin 3 x \sin 2 x) \sin x$

$\int \sin 4 x \sin 8 x d x=\frac{1}{2} \int 2 \sin 4 x \sin 8 x d x$

$\int \frac{1-\cos x}{1+\cos x} d x=\int \frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}} d x=\int \tan ^2 \frac{x}{2} d x$

$\int \frac{\cos x}{1+\cos x} d x$

Adding and subtracting 1 in the numerator of integrand,

Find the integrals of the functions in Exercises 10 to 18:

$\int \sin ^4 x d x=\int\left(\sin ^2 x\right)^2 d x=\int\left(\frac{1-\cos 2 x}{2}\right)^2 d x$

$\int \cos ^4 2 x d x=\int\left(\cos ^2 2 x\right)^2 d x$

$\int \frac{\sin ^2 x}{1+\cos x} d x=\int \frac{1-\cos ^2 x}{1+\cos x} d x=\int \frac{(1-\cos x)(1+\cos x)}{1+\cos x} d x$

Note: It may be noted that letters $a, b, c, d, \ldots, q$ of English Alphabet and letters $\alpha, \beta, \gamma, \delta$ of Greek Alphabet are generally treated as constants.

$\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x=\int \frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha} d x$

Remark: $\int \sin a d x=\sin a \int 1 d x=x \sin a$.

Please note that $\int \sin a d x \neq-\cos a$.

Let $\mathrm{I}=\int \frac{\cos x-\sin x}{1+\sin 2 x} d x=\int \frac{\cos x-\sin x}{\cos ^2 x+\sin ^2 x+2 \sin x \cos x} d x$

Put $\boldsymbol{\operatorname { c o s }} \boldsymbol{x} \boldsymbol{+} \boldsymbol{\operatorname { s i n }} \boldsymbol{x} \boldsymbol{=} \boldsymbol{t} \boldsymbol{.}$

$\therefore-\sin x+\cos x=\frac{d t}{d x}$. Therefore $(\cos x-\sin x) d x=d t$.

Therefore, from (i), I $=\int \frac{d t}{t^2}=\int t^{-2} d t=\frac{t^{-1}}{-1}+c$

$\mathrm{I}=\frac{-1}{t}+c=\frac{-1}{\cos x+\sin x}+c$.

Let $\mathrm{I}=\int \tan ^3 2 x \sec 2 x d x=\int \tan ^2 2 x \tan 2 x \sec 2 x d x$

Put $\sec 2 x=t$. Therefore $\sec 2 x \tan 2 x \frac{d}{d x}(2 x)=\frac{d t}{d x}$

$\therefore \quad 2 \sec 2 x \tan 2 x d x=d t$

Putting $t=\sec 2 x,=\frac{1}{6} \sec ^3 2 x-\frac{1}{2} \sec 2 x+c$.

$\int \tan ^4 x d x=\int \tan ^2 x \tan ^2 x d x=\int \tan ^2 x\left(\sec ^2 x-1\right) d x$

$=\int\left(\tan ^2 x \sec ^2 x-\tan ^2 x\right) d x=\int \tan ^2 x \sec ^2 x d x-\int \tan ^2 x d x$

$=\int \tan ^2 x \sec ^2 x d x-\int\left(\sec ^2 x-1\right) d x$

$=\int \tan ^2 x \sec ^2 x d x-\int \sec ^2 x d x+\int 1 d x$

↓

For this integral, put $\boldsymbol{\operatorname { t a n }} \boldsymbol{x}=\boldsymbol{t}$.

$\therefore \quad \sec ^2 x=\frac{d t}{d x} \quad$ or $\quad \sec ^2 x d x=d t$

Put $t=\tan x,=\frac{1}{3} \tan ^3 x-\tan x+x+c$.

$\int \frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x} d x=\int\left(\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}+\frac{\cos ^3 x}{\sin ^2 x \cos ^2 x}\right) d x$

$\int \frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x} d x=\int \frac{\left(1-2 \sin ^2 x\right)+2 \sin ^2 x}{\cos ^2 x} d x$

Integrate the functions in Exercises 19 to 22:

Note: Method to evaluate $\int \frac{1}{\sin ^p x \cos ^q x} d x$ if $(p+q)$ is a negative even integer ( $=-n$ (say)); then multiply Numerator and Denominator of integrand by $\sec ^n x$.

Let $\mathrm{I}=\int \frac{1}{\sin x \cos ^3 x} d x$

Here $p+q=-1-3=-4$ is a negative even integer.

So multiplying both Numerator and Denominator of integrand of (i) by $\sec ^4 x$,

Put $\tan x=t$

Putting $t=\tan x,=\log |\tan x|+\frac{1}{2} \tan ^2 x+c$.

Let $\mathrm{I}=\int \frac{\cos 2 x}{(\cos x+\sin x)^2} d x=\int \frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2} d x$

Put DENOMINATOR $\cos x+\sin x=t$

Note: Another method to evaluate integral (i) is, apply

$\int \sin ^{-1}(\cos x) d x=\int \sin ^{-1} \sin \left(\frac{\pi}{2}-x\right) d x$

Let $\mathrm{I}=\int \frac{1}{\cos (x-a) \cos (x-b)} d x$

Here $(x-a)-(x-b)=x-a-x+b=b-a$

By looking at Equation (ii), dividing and multiplying the integrand in

(i) by $\sin (b-a)$,

For Exercises 23 and 24 — choose the correct option:

$\int \frac{\sin ^2 x-\cos ^2 x}{\sin ^2 x \cos ^2 x} d x$

Let $\mathrm{I}=\int \frac{e^x(1+x)}{\cos ^2\left(e^x x\right)} d x$

Put $e^x \cdot x=t$

[We now evaluate $\int$ (T-function or Inverse T-function $\left.f(x)\right) f^{\prime}(x) d x$, put $f(x)=t]$

Applying Product Rule, $e^x \cdot 1+x e^x=\frac{d t}{d x}$

or $e^x(1+x) d x=d t$

Therefore, from (i), I $=\int \frac{d t}{\cos ^2 t}=\int \sec ^2 t d t$

$=\tan t+\mathrm{C}=\tan \left(x e^x\right)+\mathrm{C} \therefore$ Option (B) is the correct answer.

Let $\mathrm{I}=\int \frac{3 x^2}{x^6+1} d x=\int \frac{3 x^2}{\left(x^3\right)^2+1^2} d x$

Put $\boldsymbol{x}^{\mathbf{3}}=\boldsymbol{t}$

$\therefore \quad 3 x^2=\frac{d t}{d x} \quad \Rightarrow \quad 3 x^2 d x=d t$

∴ From $(i), \mathrm{I}=\int \frac{d t}{t^2+1^2}=\frac{1}{1} \tan ^{-1} \frac{t}{1}+\mathrm{C}$

Putting $t=x^3 ;=\tan ^{-1}\left(x^3\right)+\mathrm{C}$.

Note: $a x^2+b(a \neq 0)$ is called a pure quadratic.

Let $\mathrm{I}=\int \frac{1}{\sqrt{1+4 x^2}} d x=\int \frac{1}{\sqrt{(2 x)^2+1^2}} d x$

Using $\int \frac{1}{\sqrt{x^2+a^2}} d x=\log \left|x+\sqrt{x^2+a^2}\right|$,

$\mathrm{I}=\frac{\log \left|(2 x)+\sqrt{(2 x)^2+1^2}\right|}{2 \rightarrow \text { Coeff. of } x}+\mathrm{C}=\frac{1}{2} \log \left|2 x+\sqrt{4 x^2+1}\right|+\mathrm{C}$.

Let $\mathrm{I}=\int \frac{1}{\sqrt{(2-x)^2+1}} d x=\int \frac{1}{\sqrt{(2-x)^2+1}} d x$

Using $\int \frac{1}{\sqrt{x^2+a^2}} d x=\log \left|x+\sqrt{x^2+a^2}\right|$,

$=\frac{\log \left|(2-x)+\sqrt{(2-x)^2+1^2}\right|}{-1 \rightarrow \text { Coeff. of } x}+\mathrm{C}$

$=-\log \left|2-x+\sqrt{4+x^2-4 x+1}\right|+\mathrm{C}$

$=\log \left|\frac{1}{2-x+\sqrt{x^2-4 x+5}}\right|+\mathrm{C}$.

Let $\mathrm{I}=\int \frac{1}{\sqrt{9-25 x^2}} d x=\int \frac{1}{\sqrt{3^2-(5 x)^2}} d x$

Let $\mathrm{I}=\int \frac{3 x}{1+2 x^4} d x=\frac{3}{2} \int \frac{2 x}{1+2\left(x^2\right)^2} d x$

Put $x^2=t . \quad \therefore \quad 2 x=\frac{d t}{d x} \quad \Rightarrow \quad 2 x d x=d t$

Therefore, from (i), I $=\frac{3}{2} \int \frac{d t}{1+2 t^2}=\frac{3}{2} \int \frac{1}{(\sqrt{2} t)^2+1^2} d t$

Putting $t=x^2,=\frac{3}{2 \sqrt{2}} \tan ^{-1}\left(\sqrt{2} x^2\right)+\mathrm{C}$.

Let $\mathrm{I}=\int \frac{x^2}{1-x^6} d x=\int \frac{x^2}{1-\left(x^3\right)^2} d x=\frac{1}{3} \int \frac{3 x^2}{1-\left(x^3\right)^2} d x$

Put $x^3=t$. Therefore $3 x^2=\frac{d t}{d x} \Rightarrow 3 x^2 d x=d t$.

$\therefore \quad \mathrm{I}=\frac{1}{3} \int \frac{d t}{1-t^2}=\frac{1}{3} \int \frac{1}{1^2-t^2} d t=\frac{1}{3} \frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|+\mathrm{C} \left[\because \int \frac{1}{a^2-x^2} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\right]$

Putting $t=x^3,=\frac{1}{6} \log \left|\frac{1+x^3}{1-x^3}\right|+\mathrm{C}$.

Let $\mathrm{I}=\int \frac{x-1}{\sqrt{x^2-1}} d x=\int\left(\frac{x}{\sqrt{x^2-1}}-\frac{1}{\sqrt{x^2-1}}\right) d x$

Let $\mathrm{I}_1=\int \frac{2 x}{\sqrt{x^2-1}} d x$

Put $x^2-1=t$. Therefore $2 x=\frac{d t}{d x} \Rightarrow 2 x d x=d t$

Substituting this value of $\mathrm{I}_1=\int \frac{2 x}{\sqrt{x^2-1}} d x$ back into (i),

Let $\mathrm{I}=\int \frac{x^2}{\sqrt{x^6+a^6}} d x=\frac{1}{3} \int \frac{3 x^2}{\sqrt{\left(x^3\right)^2+a^6}} d x$

Put $\boldsymbol{x}^3=\boldsymbol{t}$. Therefore $3 x^2=\frac{d t}{d x} \Rightarrow 3 x^2 d x=d t$.

Therefore, from (i), $\mathrm{I}=\frac{1}{3} \int \frac{d t}{\sqrt{t^2+a^6}}=\frac{1}{3} \int \frac{1}{\sqrt{t^2+\left(a^3\right)^2}} d t$

$=\frac{1}{3} \log \left|t+\sqrt{t^2+\left(a^3\right)^2}\right|+\mathrm{C}\left[\because \int \frac{1}{\sqrt{x^2+a^2}} d x=\log \left|x+\sqrt{x^2+a^2}\right|\right]$

Putting $t=x^3,=\frac{1}{3} \log \left|x^3+\sqrt{x^6+a^6}\right|+\mathrm{C}$.

Let $\mathrm{I}=\int \frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}} d x$

Putting $t=\tan x, \mathrm{I}=\log \left|\tan x+\sqrt{\tan ^2 x+4}\right|+\mathrm{C}$.

Integrate the following functions in Exercises 10 to 18:

Note: Rule for evaluating

Write Quadratic. Take coefficient of $x^2$ common to make it unity. Then complete squares by adding and subtracting $\left(\frac{1}{2} \text { coefficient of } x\right)^2$

$\int \frac{1}{\sqrt{x^2+2 x+2}} d x=\int \frac{1}{\sqrt{x^2+2 x+1+1}} d x=\int \frac{1}{\sqrt{(x+1)^2+1^2}} d x$

Let $\mathrm{I}=\int \frac{1}{9 x^2+6 x+5} d x$

$\int \frac{1}{\text { Quadratic }} d x$

Here Quadratic expression $=9 x^2+6 x+5$

Making coefficient of $x^2$ unity, $=9\left(x^2+\frac{6 x}{9}+\frac{5}{9}\right)$

$=9\left(x^2+\frac{2 x}{3}+\frac{5}{9}\right)$

Completing the square by adding and subtracting $\left(\frac{1}{2} \text { Coefficient of } x\right)^2$

Substituting this value back into (i), $\quad \mathrm{I}=\int \frac{1}{9\left[\left(x+\frac{1}{3}\right)^2+\left(\frac{2}{3}\right)^2\right]} d x$

Let $\mathrm{I}=\int \frac{1}{\sqrt{7-6 x-x^2}} d x \quad \ldots$ (i) $\quad$ Type $\int \frac{1}{\text { Quadratic }} d x$

Here Quadratic expression is $7-6 x-x^2=-x^2-6 x+7$.

Making coefficient of $x^2$ unity, $=-\left(x^2+6 x-7\right)$.

Completing the square by adding and subtracting $\left(\frac{1}{2} \text { coefficient of } x\right)^2$

(Note: Must adjust negative sign outside Equation (ii) in the bracket as shown above because otherwise we shall get $\sqrt{-1}=i$ on taking square roots.]

Substituting the value of quadratic expression from (iii) back into (i),

Let $\mathrm{I}=\int \frac{1}{\sqrt{(x-1)(x-2)}} d x=\int \frac{1}{\sqrt{x^2-2 x-x+2}} d x$

Here quadratic expression is $x^2-3 x+2$. Coefficient of $x^2$ is already unity. Completing the square by adding and subtracting

Substituting this value back into (i), $\mathrm{I}=\int \frac{1}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}} d x$

Let $\mathrm{I}=\int \frac{1}{\sqrt{8+3 x-x^2}} d x$

Here quadratic expression is $8+3 x-x^2=-x^2+3 x+8$.

Making coefficient of $x^2$ unity, $=-\left(x^2-3 x-8\right)$.

Completing the square by adding and subtracting

$\left(\frac{1}{2} \text { coefficient of } x\right)^2=\left(\frac{3}{2}\right)^2$

(See Note given in the solution of Q.N. 12)

Let $\mathrm{I}=\int \frac{1}{\sqrt{(x-a)(x-b)}} d x=\int \frac{1}{\sqrt{x^2-b x-a x+a b}} d x$

Here Quadratic expression $=x^2-x(a+b)+a b$

Adding and subtracting $\left(\frac{1}{2} \text { coefficient of } x\right)^2=\left(\frac{a+b}{2}\right)^2$

Substituting this value back into (i),

Note: Method to evaluate $\int \frac{\text { Linear }}{\text { Quadratic }} d x$ or $\int \frac{\text { Linear }}{\sqrt{\text { Quadratic }}} d x$ or $\int$ Linear $\sqrt{\text { Quadratic }} \boldsymbol{d} \boldsymbol{x}$.

Write linear $=\mathrm{A} \frac{d}{d x}($ Quadratic $)+\mathrm{B}$.

Find values of A and B by comparing coefficients of $x$ and constant terms on both sides.

Let $\mathrm{I}=\int \frac{4 x+1}{\sqrt{2 x^2+x-3}} d x$

Here $\frac{d}{d x}$ (Quadratic $2 x^2+x-3$ ) is ( $4 x+1$ ), the numerator.

So put $2 x^2+x-3=t$.

Let $\mathrm{I}=\int \frac{x+2}{\sqrt{x^2-1}} d x=\int\left(\frac{x}{\sqrt{x^2-1}}+\frac{2}{\sqrt{x^2-1}}\right) d x$

Let $\mathrm{I}_1=\int \frac{x}{\sqrt{x^2-1}} d x=\frac{1}{2} \int \frac{2 x}{\sqrt{x^2-1}} d x$

Put $x^2-1=t$. Therefore $2 x=\frac{d t}{d x} \quad$ or $\quad 2 x d x=d t$

Substituting the value of $\left(\mathrm{I}_1=\right) \int \frac{x}{\sqrt{x^2-1}} d x=\sqrt{x^2-1}$ into (i)

Let $\mathrm{I}=\int \frac{5 x-2}{1+2 x+3 x^2} d x$

Let Linear $=\mathbf{A} \frac{\boldsymbol{d}}{\boldsymbol{d} \boldsymbol{x}}($ Quadratic $)+\mathbf{B}$

i.e., $\quad 5 x-2=\mathrm{A} \frac{d}{d x}\left(1+2 x+3 x^2\right)+\mathrm{B}$

or $\quad 5 x-2=\mathrm{A}(2+6 x)+\mathrm{B}$

i.e., $\quad 5 x-2=2 \mathrm{~A}+6 \mathrm{~A} x+\mathrm{B}$

Equating coefficients of $x, 6 \mathrm{~A}=5 \Rightarrow \mathrm{~A}=\frac{5}{6}$

Matching constant terms, $2 \mathrm{~A}+\mathrm{B}=-2$

Putting $\mathrm{A}=\frac{5}{6}, \frac{10}{6}+\mathrm{B}=-2$

$\Rightarrow \quad \mathrm{B}=-2-\frac{10}{6}=\frac{-22}{6} \quad$ or $\quad \mathrm{B}=\frac{-11}{3}$

Substituting values of A and B back into (ii), $5 x-2=\frac{5}{6}(2+6 x)-\frac{11}{3}$

Substituting this value of $5 x-2$ back into (i),

Here $\mathrm{I}_1=\int \frac{2+6 x}{1+2 x+3 x^2} d x$

Put Denominator $1+2 x+3 x^2=t$.

Again $\left.\mathrm{I}_2=\int \frac{1}{1+2 x+3 x^2} d x=\int \frac{1}{3 x^2+2 x+1} d x \right\rvert\, \int \frac{1}{\text { Quadratic }} d x$

Now Quadratic Expression $=3 x^2+2 x+1$.

Making coefficient of $x^2$ unity $=3\left(x^2+\frac{2}{3} x+\frac{1}{3}\right)$

Substituting values of $\mathrm{I}_1$ and $\mathrm{I}_2$ from (iv) and (v) back into (iii), we have

$\mathrm{I}=\frac{5}{6} \log \left|1+2 x+3 x^2\right|-\frac{11}{3} \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+c$.

Integrate the functions in Exercises 19 to 23:

Let $\mathrm{I}=\int \frac{6 x+7}{\sqrt{(x-5)(x-4)}} d x=\int \frac{6 x+7}{\sqrt{x^2-4 x-5 x+20}} d x$

Let Linear $=\mathbf{A} \frac{\boldsymbol{d}}{\boldsymbol{d} \boldsymbol{x}}($ Quadratic $)+\mathbf{B}$

Equating coefficients of $x, 2 \mathrm{~A}=6 \Rightarrow \mathrm{~A}=3$

Matching constant terms, $-9 \mathrm{~A}+\mathrm{B}=7$.

Putting $\mathrm{A}=3,-27+\mathrm{B}=7 \Rightarrow \mathrm{~B}=34$

Substituting values of A and B back into (ii),

Substituting this value of $6 x+7$ back into (i),

Put $x^2-9 x+20=t . \quad \therefore \quad 2 x-9=\frac{d t}{d x}$

Substituting values of $\mathrm{I}_1$ and $\mathrm{I}_2$ from (iv) and (v) back into (iii),

Let $\mathrm{I}=\int \frac{x+2}{\sqrt{4 x-x^2}} d x$

Let Linear $=\mathrm{A} \frac{d}{d x}($ Quadratic $)+\mathrm{B}$

i.e., $\quad x+2=\mathrm{A}(4-2 x)+\mathrm{B}$

Equating coefficients of $x:-2 \mathrm{~A}=1 \Rightarrow \mathrm{~A}=\frac{-1}{2}$

Equating constants: $4 \mathrm{~A}+\mathrm{B}=2$

Putting $\mathrm{A}=\frac{-1}{2}, \quad-2+\mathrm{B}=2 \Rightarrow \mathrm{~B}=4$

Substituting values of A and B back into (ii), $x+2=\frac{-1}{2}(4-2 x)+4$

Substituting this value of $x+2$ back into (i),

Put $4 x-x^2=t \quad \therefore \quad 4-2 x=\frac{d t}{d x} \quad \Rightarrow \quad(4-2 x) d x=d t$

Quadratic Expression is $4 x-x^2=-x^2+4 x$

Substituting values of $\mathrm{I}_1$ and $\mathrm{I}_2$ from (iv) and (v) back into (iii),

Let $\mathrm{I}=\int \frac{x+2}{\sqrt{x^2+2 x+3}} d x$

Let Linear $=\mathbf{A} \frac{\boldsymbol{d}}{\boldsymbol{d} \boldsymbol{x}}($ Quadratic $)+\mathbf{B}$

Equating coefficients of $x, 2 \mathrm{~A}=1 \Rightarrow \mathrm{~A}=\frac{1}{2}$

Matching constant terms, $2 \mathrm{~A}+\mathrm{B}=2$

Putting $\mathrm{A}=\frac{1}{2}, 1+\mathrm{B}=2 \Rightarrow \mathrm{~B}=1$

Substituting values of A and B back into (ii), $x+2=\frac{1}{2}(2 x+2)+1$

Substituting this value of $(x+2)$ back into (i),

Substituting values from (iv) and (v) back into (iii),

Let $\mathrm{I}=\int \frac{x+3}{x^2-2 x-5} d x$

Let $x+3=\mathrm{A} \frac{d}{d x}\left(x^2-2 x-5\right)+\mathrm{B}$

or $x+3=\mathrm{A}(2 x-2)+\mathrm{B}$

Equating coefficients of $x$ on both sides, $2 \mathrm{~A}=1 \Rightarrow \mathrm{~A}=\frac{1}{2}$

Matching constant terms, $-2 \mathrm{~A}+\mathrm{B}=3$

Putting $\mathrm{A}=\frac{1}{2},-1+\mathrm{B}=3 \Rightarrow \mathrm{~B}=4$

Substituting values of A and B back into (ii), $x+3=\frac{1}{2}(2 x-2)+4$

Substituting this value back into (i),

Put $\quad x^2-2 x-5=t$. Therefore $(2 x-2)=\frac{d t}{d x} \Rightarrow(2 x-2) d x=d t$

Substituting values of $\mathrm{I}_1$ and $\mathrm{I}_2$ from (iv) and (v) back into (iii),

Let $\mathrm{I}=\int \frac{5 x+3}{\sqrt{x^2+4 x+10}} d x$

Let Linear $=\mathbf{A} \frac{\boldsymbol{d}}{\boldsymbol{d} \boldsymbol{x}}($ Quadratic $)+\mathbf{B}$

i.e., $\quad 5 x+3=\mathrm{A}(2 x+4)+\mathrm{B}$

Equating coefficients of $x$ on both sides, $2 \mathrm{~A}=5 \Rightarrow \mathrm{~A}=\frac{5}{2}$

Matching constant terms, $4 \mathrm{~A}+\mathrm{B}=3$

Putting $\mathrm{A}=\frac{5}{2}, 10+\mathrm{B}=3 \quad \Rightarrow \mathrm{~B}=-7$

Substituting values of A and B back into (ii), $5 x+3=\frac{5}{2}(2 x+4)-7$

Putting this value in $(i), \mathrm{I}=\int \frac{\frac{5}{2}(2 x+4)-7}{\sqrt{x^2+4 x+10}} d x$

or $\quad \mathrm{I}=\frac{5}{2} \mathrm{I}_1-7 \mathrm{I}_2$

Put $x^2+4 x+10=t$. Therefore $2 x+4=\frac{d t}{d x} \Rightarrow(2 x+4) d x=d t$

Substituting values of $\mathrm{I}_1$ and $\mathrm{I}_2$ from (iv) and (v) back into (iii),

Choose the correct option in Exercises 24 and 25.

$\int \frac{d x}{x^2+2 x+2}=\int \frac{1}{x^2+2 x+1+1} d x=\int \frac{1}{(x+1)^2+1^2} d x$

Hence, Option (B) is the correct answer.

Let $\mathrm{I}=\int \frac{d x}{\sqrt{9 x-4 x^2}}=\int \frac{d x}{\sqrt{-4 x^2+9 x}}$

Here Quadratic expression is $-4 x^2+9 x=-4\left(x^2-\frac{9}{4} x\right)$

Substituting this value back into (i),

We integrate the rational function $\frac{x}{(x+1)(x+2)}$.

Let integrand $\frac{x}{(x+1)(x+2)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+2}$

(Partial Fraction Decomposition)

Multiplying both sides by the L.C.M. $=(x+1)(x+2)$,

Equating coefficients of $x$ on both sides, $\mathrm{A}+\mathrm{B}=1$

Matching constant terms, $2 \mathrm{~A}+\mathrm{B}=0$

We now solve Equations (ii) and (iii) to determine A and B .

Equation (iii) – Equation (ii) gives, $\mathrm{A}=-1$

Substituting $\mathrm{A}=-1$ back into (ii), $-1+\mathrm{B}=1 \Rightarrow \mathrm{~B}=2$

Substituting values of A and B back into (i), $\frac{x}{(x+1)(x+2)}=\frac{-1}{x+1}+\frac{2}{x+2}$

We integrate the rational function $\frac{1}{x^2-9}$

OR

Integrand $\frac{1}{x^2-9}=\frac{1}{(x-3)(x+3)}=\frac{\mathrm{A}}{x-3}+\frac{\mathrm{B}}{x+3}$

Now proceed as in the solution of Q.No.1.

We integrate the rational function $\frac{3 x-1}{(x-1)(x-2)(x-3)}$

Let integrand $\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x-3}$

Multiplying both sides by the L.C.M. $=(x-1)(x-2)(x-3)$, we have

Equating coefficients of $x^2, x$ and constant terms on both sides,

Coefficients of $\boldsymbol{x}^{\mathbf{2}}: \mathrm{A}+\mathrm{B}+\mathrm{C}=0$

Coefficient of $\boldsymbol{x}:-5 \mathrm{~A}-4 \mathrm{~B}-3 \mathrm{C}=3$ or $5 \mathrm{~A}+4 \mathrm{~B}+3 \mathrm{C}=-3$

Constants: $6 \mathrm{~A}+3 \mathrm{~B}+2 \mathrm{C}=-1$

We now solve (ii), (iii) and (iv) to determine A, B, C.

We build two equations in two unknowns A and B.

Equation (iii) – 3 Equation (i) gives (to eliminate C),

Equation (iv) -2 Equation (i) gives (to eliminate C),

Equation (vi) – Equation (v) gives (to eliminate B),

Substituting $\mathrm{A}=1$ back into (v), $2+\mathrm{B}=-3 \Rightarrow \mathrm{~B}=-5$

Substituting $\mathrm{A}=1$ and $\mathrm{B}=-5$ back into (ii), $1-5+\mathrm{C}=0$

or $\mathrm{C}-4=0$ or $\mathrm{C}=4$

Substituting values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ back into (i),

We integrate the rational function $\frac{x}{(x-1)(x-2)(x-3)}$.

Let integrand $\frac{x}{(x-1)(x-2)(x-3)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x-3}$

(Partial Fraction Decomposition)

Multiplying both sides by the L.C.M. $=(x-1)(x-2)(x-3)$,

Equating coefficients of $x^2, x$ and constant terms on both sides,

Constants: $6 \mathrm{~A}+3 \mathrm{~B}+2 \mathrm{C}=0$

We now solve Equations (ii), (iii) and (iv) to determine A, B, C.

We build two equations in two unknowns A and B.

Equation (iii) $-3 \times$ Equation (ii) gives | To eliminate C

Equation (iv) $-2 \times$ Equation (ii) gives To eliminate C

Equation (vi) – Equation (v) gives (To eliminate B)

Putting $\mathrm{A}=\frac{1}{2}$ in $(v), 1+\mathrm{B}=-1 \Rightarrow \mathrm{~B}=-2$

Substituting $\mathrm{A}=\frac{1}{2}$ and $\mathrm{B}=-2$ back into (ii),

Substituting these values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ back into (i), we have

We integrate the rational function $\frac{2 x}{x^2+3 x+2}$.

Now $x^2+3 x+2=x^2+2 x+x+2=x(x+2)+1(x+2)$

(Partial Fraction Decomposition)

Multiplying both sides by L.C.M. $=(x+1)(x+2)$,

Equating coefficients of $x$ and constant terms on both sides,

Coefficients of $\boldsymbol{x}: \mathrm{A}+\mathrm{B}=2$

Constant terms: $2 \mathrm{~A}+\mathrm{B}=0$

We now solve (ii) and (iii) to determine A and B .

(iii) – (ii) gives $\mathrm{A}=-2$.

Substituting $\mathrm{A}=-2$ back into (ii), $-2+\mathrm{B}=2$.

Substituting values of A and B back into (i), $\frac{2 x}{x^2+3 x+2}=\frac{-2}{x+1}+\frac{4}{x+2}$

Remark: Alternative method to evaluate $\int \frac{2 x}{x^2+3 x+2} d x$

is $\int \frac{\text { Linear }}{\text { Quadratic }} d x$ as explained in solutions in Exercise 7.4

(Exercise 18 and Exercise 22.

To integrate (rational) function $\frac{1-x^2}{x(1-2 x)}=\frac{1-x^2}{x-2 x^2}=\frac{-x^2+1}{-2 x^2+x}$

[Here Degree of numerator $=$ Degree of Denominator $=2$

∴ We must divide numerator by denominator to make the degree of numerator smaller than degree of denominator so that we can form partial fractions.]

Let integrand $\frac{-\frac{x}{2}+1}{x(1-2 x)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{1-2 x}$

Multiplying both sides by the L.C.M. $=x(1-2 x)$,

Equating coefficients of $x,-2 \mathrm{~A}+\mathrm{B}=\frac{-1}{2}$

Matching constant terms, $\mathrm{A}=1$

Substituting $\mathrm{A}=1$ from (iv) back into (iii),

Substituting values of A and B back into (ii),

Substituting this value back into (i),

Integrate the following functions in Exercises 7 to 12:

We integrate the rational function $\frac{x}{\left(x^2+1\right)(x-1)}$.

Let integrand $\frac{x}{\left(x^2+1\right)(x-1)}=\frac{\mathrm{A} x+\mathrm{B}}{x^2+1}+\frac{\mathrm{C}}{x-1}$

(Partial Fraction Decomposition)

Multiplying both sides by the L.C.M. $=\left(x^2+1\right)(x-1)$ on both sides,

Equating coefficients of $x^2, x$ and constant terms on both sides,

Let us solve Equations (ii), (iii) and (iv) for A, B, C Adding (ii) and (iii) to eliminate $\mathrm{A}, \mathrm{B}+\mathrm{C}=1$

Adding (iv) and (v), $2 \mathrm{C}=1 \Rightarrow \mathrm{C}=\frac{1}{2}$

From (iv), $-\mathrm{B}=-\mathrm{C} \Rightarrow \mathrm{B}=\mathrm{C}=\frac{1}{2}$

From (ii), $\quad \mathrm{A}=-\mathrm{C}=\frac{-1}{2}$

Substituting these values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ back into (i),

We integrate the rational function $\frac{x}{(x-1)^2(x+2)}$.

Let integrand $\frac{x}{(x-1)^2(x+2)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{(x-1)^2}+\frac{\mathrm{C}}{x+2}$

(Partial Fraction Decomposition)

Multiplying both sides of (i) by L.C.M. $=(x-1)^2(x+2)$

or $x=\mathrm{A}\left(x^2+2 x-x-2\right)+\mathrm{B}(x+2)+\mathrm{C}\left(x^2+1-2 x\right)$

or $x=\mathrm{A} x^2+\mathrm{A} x-2 \mathrm{~A}+\mathrm{B} x+2 \mathrm{~B}+\mathrm{C} x^2+\mathrm{C}-2 \mathrm{C} x$

Equating coefficients of $x^2, x$ and constant terms on both sides

$x^2$

$\boldsymbol{x}$

Constants $-2 \mathrm{~A}+2 \mathrm{~B}+\mathrm{C}=0$

Let us solve (ii), (iii) and (iv) for A, B, C

From (ii), $\quad \mathrm{A}=-\mathrm{C}$

Substituting $\mathrm{A}=-\mathrm{C}$ back into (iv), $\quad 2 \mathrm{C}+2 \mathrm{~B}+\mathrm{C}=0$

Substituting values of A and B back into (iii),

Substituting $\mathrm{C}=\frac{-2}{9}, \mathrm{~B}=\frac{-3 \mathrm{C}}{2}=\frac{-3}{2}\left(\frac{-2}{9}\right)=\frac{1}{3} \therefore \mathrm{~A}=-\mathrm{C}=\frac{2}{9}$ Putting these values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ back into (i),

We integrate the rational function $\frac{3 x+5}{x^3-x^2-x+1}$.

Now denominator $=x^3-x^2-x+1$

∴ Integrand $\frac{3 x+5}{x^3-x^2-x+1}=\frac{3 x+5}{(x-1)^2(x+1)}$

Multiplying both sides by the L.C.M. $=(x-1)^2(x+1)$,

Equating coefficients of $x^2, x$ and constant terms on both sides,

$x^2$

$x$

Constants

We now solve Equations (ii), (iii) and (iv) to determine A, B, C.

From (ii), $\mathrm{A}=-\mathrm{C}$ and from (iii), $\mathrm{B}=2 \mathrm{C}+3$

Substituting these values of A and B back into (iv),

Putting these values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ in (i)

To integrate the rational function $\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}$.

Let integrand $\frac{2 x-3}{\left(x^2-1\right)(2 x+3)}=\frac{2 x-3}{(x-1)(x+1)(2 x+3)}$

Multiplying both sides by L.C.M. $=(x-1)(x+1)(2 x+3)$,

Equating coefficients of $x^2, x$ and constant terms on both sides,

We now solve Equations (ii), (iii) and (iv) to determine A, B, C.

Equation (ii) + Equation (iv) gives (to eliminate C)

Adding Equations (iii) and (v), $10 \mathrm{~A}=-1 \Rightarrow \mathrm{~A}=\frac{-1}{10}$

Substituting $\mathrm{A}=\frac{-1}{10}$ back into (iii), $\frac{-5}{10}+\mathrm{B}=2 \Rightarrow \mathrm{~B}=2+\frac{1}{2}=\frac{5}{2}$

Substituting values of A and B back into (ii),

Substituting values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ back into (i),

To integrate the rational function $\frac{5 x}{(x+1)\left(x^2-4\right)}$.

Let integrand $\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{5 x}{(x+1)(x+2)(x-2)}$

…(i) (Partial Fraction Decomposition)

Multiplying both sides of (i) by L.C.M.

Equating coefficients of $x^2, x$ and constant terms on both sides,

$\boldsymbol{x}^{\mathbf{2}}$

$\boldsymbol{x}$

$-\mathrm{B}+3 \mathrm{C}=5$

Constants $-4 \mathrm{~A}-2 \mathrm{~B}+2 \mathrm{C}=0$

Dividing by $-2,2 \mathrm{~A}+\mathrm{B}-\mathrm{C}=0$

Let us solve (ii), (iii) and (iv) for A, B, C

Equation (ii) $\times 2-$ Equation (iv) gives (To eliminate A) because Equation (iii) does not involve A.

Adding Equations (iii) and (v),

Substituting $\mathrm{C}=\frac{5}{6}$ back into (iii), $-\mathrm{B}+\frac{15}{6}=5 \Rightarrow-\mathrm{B}=5-\frac{15}{6}$

Substituting $\mathrm{B}=\frac{-5}{2}$ and $\mathrm{C}=\frac{5}{6}$ back into (ii), $\mathrm{A}-\frac{5}{2}+\frac{5}{6}=0$

Substituting values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ back into (i),

Here degree of numerator is greater than degree of denominator.

Therefore, dividing the numerator by the denominator,

Multiplying both sides by the L.C.M. $=(x+1)(x-1)$, we have

By equating the coefficients of $x$ and constant terms, we get

(iii) + (iv) gives $2 \mathrm{~B}=3 \Rightarrow \mathrm{~B}=\frac{3}{2}$

Substituting $\mathrm{B}=\frac{3}{2}$ back into (iii), we get $\mathrm{A}+\frac{3}{2}=2$ or $\mathrm{A}=\frac{1}{2}$

Putting values of A and B in eqn. (ii), we have

Substituting this value of $\frac{2 x+1}{x^2-1}$ back into (i),

Integrate the following functions in Exercises 13 to 17:

To find integral of the Rational function $\frac{2}{(1-x)\left(1+x^2\right)}$.

(Partial Fraction Decomposition)

Multiplying both sides by the L.C.M. $=(1-x)\left(1+x^2\right)$

Equating coefficients of $x^2, x$ and constant terms, we have

Constant terms $\mathrm{A}+\mathrm{C}=2$

Let us solve (ii), (iii), (iv) for A, B, C

From (ii), $\quad \mathrm{A}=\mathrm{B}$ and from (iii), $\quad \mathrm{B}=\mathrm{C}$

Substituting $\mathrm{A}=\mathrm{C}$ back into (iv), $\mathrm{C}+\mathrm{C}=2$ or $2 \mathrm{C}=2$ or $\mathrm{C}=1$

Substituting these values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ back into (i),

Note: $\log |1-x|=\log |-(x-1)|$

To find integral of rational function $\frac{3 x-1}{(x+2)^2}$.

Form $\int \frac{\text { Polynomial function }}{\text { (Linear) }^k} d x$ where $k$ is a positive integer, put Linear $=t$.

Here put $x+2=t$

Substituting these values back into (i),

Putting $t=x+2,=3 \log |x+2|+\frac{7}{x+2}+c$.

Remark: Alternative solution is Let $\frac{3 x-1}{(x+2)^2}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B}}{(x+2)^2}$.

To find integral of $\frac{1}{x^4-1}$.

Let integrand $\frac{1}{x^4-1}=\frac{1}{\left(x^2-1\right)\left(x^2+1\right)}$.

Put $x^2=y$ only to form partial fractions.

Multiplying both sides by the L.C.M. $=(y-1)(y+1)$

Equating coeffs. of $y$ and constant terms, we have

Coefficients of $\boldsymbol{y}$ : $\quad \mathrm{A}+\mathrm{B}=0$

Constant terms $\quad \mathrm{A}-\mathrm{B}=1$

Adding (ii) and (iii), $2 \mathrm{~A}=1 \quad \Rightarrow \mathrm{~A}=\frac{1}{2}$

Substituting $\mathrm{A}=\frac{1}{2}$ back into (ii), $\frac{1}{2}+\mathrm{B}=0 \Rightarrow \mathrm{~B}=\frac{-1}{2}$

Substituting values of $\mathrm{A}, \mathrm{B}$ and $\boldsymbol{y}$ back into (i),

Note: Must put $\boldsymbol{y}=\boldsymbol{x}^{\mathbf{2}}$ in (i) along with values of A and B before writing values of integrals.

Remark: Alternative solution is:

But the above given solution is better.

Let $\mathrm{I}=\int \frac{1}{x\left(x^n+1\right)} d x$

Multiplying both numerator and denominator of integrand by $n x^{n-1}$.

Put $x^n=t$. Therefore $n x^{n-1}=\frac{d t}{d x} . \quad \therefore n x^{n-1} d x=d t$.

Therefore, from (i), I $=\frac{1}{n} \int \frac{d t}{t(t+1)}=\frac{1}{n} \int \frac{1}{t(t+1)} d t$

Adding and subtracting $t$ in the numerator of integrand,

Putting $t=x^n,=\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|+c$

Remark: Alternative solution for $\int \frac{1}{t(t+1)} d t$ is:

Let $\frac{1}{t(t+1)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B}}{t+1}$.

But the above given solution is better.

Let $\mathrm{I}=\int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x$

Put $\sin x=t$. Therefore $\cos x=\frac{d t}{d x} \Rightarrow \cos x d x=d t$,

$\therefore \quad$ From $(i), \int \frac{1}{(1-t)(2-t)} d t=\int \frac{(2-t)-(1-t)}{(1-t)(2-t)} d t$

[ ∵ Difference of two factors in the denominator namely $1-t$ and $2-t$ is $(2-t)-(1-t)=2-t-1+t=1$ ]

Putting $t=\sin x,=\log \left|\frac{2-\sin x}{1-\sin x}\right|+c$

Remark: Alternative solution for $\int \frac{1}{(1-t)(2-t)} d t$ is

Let $\frac{1}{(1-t)(2-t)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{2-t}$

Integrate the following functions for Exercises 18 to 21:

To integrate the rational function $\frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}$.

Put $\boldsymbol{x}^2=\boldsymbol{y}$ in the integrand to get

Here degree of numerator $=$ degree of denominator ( $=2$ )

So have to perform long division to make the degree of numerator smaller than degree of denominator so that the concept of forming partial fractions becomes valid.

$\therefore \quad$ From (i) and (ii),

Let us form partial fractions of $\frac{(-4 y-10)}{(y+3)(y+4)}$.

Let $\frac{-4 y-10}{(y+3)(y+4)}=\frac{\mathrm{A}}{y+3}+\frac{\mathrm{B}}{y+4}$

Multiplying both sides by the L.C.M. $=(y+3)(y+4)$

Equating coefficients of $y, \quad \mathrm{~A}+\mathrm{B}=-4$

Matching constant terms, $4 \mathrm{~A}+3 \mathrm{~B}=-10$

We now solve Equations (v) and (vi) to determine A and B .

Equation (v) $\times 4$ gives, $4 \mathrm{~A}+4 \mathrm{~B}=-16$

Equation (vi) – Equation (vii) gives, $-\mathrm{B}=6$ or $\mathrm{B}=-6$.

Substituting $\mathrm{B}=-6$ back into (v), $\mathrm{A}-6=-4 \Rightarrow \mathrm{~A}=-4+6=2$

Substituting these values of A and B back into (iv),

Substituting this value back into (iii),

In R.H.S., Putting $y=x^2$ (before integration)

Let $\mathrm{I}=\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x$

Put $x^2=t$. Differentiating both sides $2 x d x=d t$

Dividing and multiplying by 2 ,

Let $\mathrm{I}=\int \frac{1}{x\left(x^4-1\right)} d x$

Multiplying both numerator and denominator of integrand by $4 x^3$.

Put $\boldsymbol{x}^4=\boldsymbol{t}$. Therefore $4 x^3=\frac{d t}{d x} \Rightarrow 4 x^3 d x=d t$.

Putting $t=x^4,=\frac{1}{4} \log \left|\frac{x^4-1}{x^4}\right|+c$.

Remark: Alternative solution is:

But the solution given above is much better.

Let $\mathrm{I}=\int \frac{1}{e^x-1} d x$

Put $\boldsymbol{e}^{\boldsymbol{x}}=\boldsymbol{t}$. Therefore $e^x=\frac{d t}{d x} \Rightarrow e^x d x=d t \Rightarrow d x=\frac{d t}{e^x}$

(Rule for evaluating $\int \boldsymbol{f}\left(\boldsymbol{e}^{\boldsymbol{x}}\right) \boldsymbol{d} \boldsymbol{x}$, put $\boldsymbol{e}^{\boldsymbol{x}}=\boldsymbol{t}$ )

Choose the correct option in each of the Exercises 22 and 23:

Let integrand $\frac{x}{(x-1)(x-2)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}$

(Partial Fraction Decomposition)

Multiplying both sides by the L.C.M. $=(x-1)(x-2)$,

Equating coefficients of $x$ and constant terms on both sides,

Coefficients of $x$ : $\quad \mathrm{A}+\mathrm{B}=1$

Constant terms: $-2 \mathrm{~A}-\mathrm{B}=0$

Let us solve (ii) and (iii) for A and B

Adding (ii) and (iii), $-\mathrm{A}=1$ or $\mathrm{A}=-1$

Putting $\mathrm{A}=-1$ in (ii) $-1+\mathrm{B}=1$ or $\mathrm{B}=2$

Substituting values of A and B back into (i),

Hence, Option (B) is the correct answer.

Let $\mathrm{I}=\int \frac{1}{x\left(x^2+1\right)} d x$

Multiplying both numerator and denominator of integrand by $2 x$.

Put $x^2=t . \quad \therefore 2 x=\frac{d t}{d x} \Rightarrow 2 x d x=d t$

Therefore, from (i), I $=\int \frac{d t}{2 t(t+1)}=\frac{1}{2} \int \frac{1}{t(t+1)} d t$

Adding and subtracting $t$ in the numerator of integrand,

Putting $t=x^2, \mathrm{I}=\frac{1}{2}\left(\log \left|x^2\right|-\log \left|x^2+1\right|\right)+c$

Hence, Option (A) is the correct answer.

$\int_{\text {I II }} x \sin x d x$

Applying Product Rule I $\int$ II $d x-\int\left(\frac{d}{d x}\right.$ (I) $\int$ II $\left.d x\right) d x$

$=x(-\cos x)-\int 1(-\cos x) d x=-x \cos x-\int-\cos x d x$

$=-x \cos x+\int \cos x d x=-x \cos x+\sin x+c$

Note: $\int \sin x d x=-\cos x$.

$\int_{\text {I }} x \sin 3 x d x$

Applying Product Rule I $\int$ II $d x-\int\left(\frac{d}{d x}(\mathrm{I}) \int\right.$ II $\left.d x\right) d x$

$=x \int \sin 3 x d x-\int\left(\frac{d}{d x}(x) \int \sin 3 x d x\right) d x$

$=x\left(\frac{-\cos 3 x}{3}\right)-\int\left[1\left(\frac{-\cos 3 x}{3}\right)\right] d x+c$

$=\frac{-1}{3} x \cos 3 x+\frac{1}{3} \int \cos 3 x d x+c$

$\int_{\text {I II }} x^2 e^x d x$

Applying Product Rule I $\int$ II $d x-\int\left(\frac{d}{d x}(\right.$ I $) \int$ II $\left.d x\right) d x$

Again Applying Product Rule

$\int x \log x d x=\int(\log x) \cdot x d x$

Applying Product Rule I $\int$ II $d x-\int\left[\frac{d}{d x}\right.$ (I) $\int$ II $\left.d x\right] d x$

$=(\log x) \int x d x-\int\left[\frac{d}{d x}(\log x) \int x d x\right] d x$

$=(\log x) \frac{x^2}{2}-\int \frac{1}{x} \frac{x^2}{2} d x=\frac{1}{2} x^2 \log x-\frac{1}{2} \int x d x$

$=\frac{1}{2} x^2 \log x-\frac{1}{2} \frac{x^2}{2}+c=\frac{x^2}{2} \log x-\frac{x^2}{4}+c$.

$\int x \log 2 x d x=\int(\log 2 x) \cdot x d x$

I II

Applying Product Rule I $\int$ II $d x-\int\left(\frac{d}{d x}(\right.$ I $) \int$ II $\left.d x\right) d x$

$=(\log 2 x) \int x d x-\int\left(\frac{d}{d x}(\log 2 x) \int x d x\right) d x$

$=(\log 2 x) \frac{x^2}{2}-\int \frac{1}{2 x} \cdot 2 \cdot \frac{x^2}{2} d x$

$\int x^2 \log x d x=\int(\log x) x^2 d x$

I II

Applying Product Rule: I $\int$ II $d x-\int\left(\frac{d}{d x}\right.$ (I) $\int$ II $\left.d x\right) d x$

Let $\mathrm{I}=\int x \sin ^{-1} x d x$.

Put $\boldsymbol{x}=\boldsymbol{\operatorname { s i n }} \theta$. Differentiating both sides $d x=\cos \theta d \theta$

Applying integration by parts

Let I $=\int x \tan ^{-1} x d x=\int\left(\tan ^{-1} x\right) \cdot x d x$

Integrate the functions in Exercises 9 to 15:

Let $\mathrm{I}=\int x \cos ^{-1} x d x$

Put $\boldsymbol{\operatorname { c o s }}^{-1} \boldsymbol{x} \boldsymbol{=} \boldsymbol{\theta}$. Therefore $\boldsymbol{x}=\boldsymbol{\operatorname { c o s }} \boldsymbol{\theta}$.

Applying Product Rule: I $\int$ II $d \theta-\int\left[\frac{d}{d \theta}(\mathrm{I}) \int \mathrm{II} d \theta\right] d \theta$

Putting $\cos \theta=x$ and $\theta=\cos ^{-1} x$;

Put $\boldsymbol{x}=\boldsymbol{\operatorname { s i n }} \theta$. Differentiating both sides, $d x=\cos \theta d \theta$

Let $\mathrm{I}=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^2}} d x$

Put $\cos ^{-1} x=\theta . \quad \Rightarrow \quad x=\cos \theta$

Therefore $\frac{d x}{d \theta}=-\sin \theta \Rightarrow d x=-\sin \theta d \theta$

Applying Product Rule: I $\int$ II $d \theta-\int\left[\frac{d}{d \theta}\right.$ (I) $\int$ II $\left.d \theta\right] d \theta$

Putting $\theta=\cos ^{-1} x$ and $\cos \theta=x$,

$\int_{\text {I II }} x \sec ^2 x d x$

Applying Product Rule: I $\int$ II $d x-\int\left[\frac{d}{d x}\right.$ (I) $\int$ II $\left.d x\right] d x$

Let $\mathrm{I}=\int \tan ^{-1} x d x=\int\left(\tan ^{-1} x\right) .1 d x$

$\int x(\log x)^2 d x=\int(\log x)^2 \cdot x d x$

Applying Product Rule: I $\int$ II $d x-\int\left[\frac{d}{d x}(\mathrm{I}) \int\right.$ II $\left.d x\right] d x$

I II

Again applying Product Rule: I $\int$ II $d x-\int\left[\frac{d}{d x}(\right.$ I $) \int$ II $\left.d x\right] d x$

$\int\left(x^2+1\right) \log x d x=\int(\log x)\left(x^2+1\right) d x$

I II

Applying Product Rule: I $\int$ II $d x-\int\left[\frac{d}{d x}\right.$ (I) $\int$ II $\left.d x\right] d x$

Integrate the functions in Exercises 16 to 22:

Here $\mathrm{I}=\int e^x(\sin x+\cos x) d x$

It is of the form $\int e^x\left[f(x)+f^{\prime}(x)\right] d x$

Let us take $f(x)=\sin x$ so that $f^{\prime}(x)=\cos x$

Here $\mathrm{I}=\int \frac{x e^x}{(x+1)^2} d x=\int \frac{(x+1)-1}{(x+1)^2} e^x d x$

It is of the form $\int e^x\left[f(x)+f^{\prime}(x)\right] d x$

Let us take $f(x)=\frac{1}{x+1}$ so that $f^{\prime}(x)=\frac{d}{d x}\left[(x+1)^{-1}\right]$

Here $\mathrm{I}=\int e^x \cdot \frac{1+\sin x}{1+\cos x} d x=\int e^x \cdot \frac{1+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}} d x$

It is of the form $\int e^x\left[f(x)+f^{\prime}(x)\right] d x$

Let us take $f(x)=\tan \frac{x}{2}$ so that $f^{\prime}(x)=\frac{1}{2} \sec ^2 \frac{x}{2}$

Let $\mathrm{I}=\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x$

It is of the form $\int e^x\left(f(x)+f^{\prime}(x)\right) d x$

Here $f(x)=\frac{1}{x}=x^{-1}$ and so $f^{\prime}(x)=(-1) x^{-2}=\frac{-1}{x^2}$

Here $\mathrm{I}=\int \frac{(x-3) e^x}{(x-1)^3} d x=\int \frac{(x-1)-2}{(x-1)^3} e^x d x$

It is of the form $\int e^x\left[f(x)+f^{\prime}(x)\right] d x$

Let us take $f(x)=\frac{1}{(x-1)^2}$ so that $f^{\prime}(x)=\frac{d}{d x}\left[(x-1)^{-2}\right]$

Note: Rule for evaluating $\int e^{\boldsymbol{a x}} \boldsymbol{\operatorname { s i n }} \boldsymbol{b} \boldsymbol{x} d \boldsymbol{x}$ or $\int \boldsymbol{e}^{\boldsymbol{a x}} \boldsymbol{\operatorname { c o s }} \boldsymbol{b} \boldsymbol{x} d \boldsymbol{x}$

Let $\mathrm{I}=\int e^{a x} \sin b x d x$ or $\int e^{a x} \cos b x d x$

I II I II

Integrate twice by product Rule and transpose term containing I from R.H.S. to L.H.S.

Let $\mathrm{I}=\underset{\text { I }}{\int} e^{2 x} \sin x d x$

Applying Product Rule: I $\int$ II $d x-\int\left[\frac{d}{d x}(\right.$ I $) \int$ II $\left.d x\right] d x$

$\Rightarrow \mathrm{I}=e^{2 x}(-\cos x)-\int e^{2 x} \cdot 2 \cdot(-\cos x) d x$

I I I

Again Applying Product Rule:

Transposing – 4 I to L.H.S.; $5 \mathrm{I}=e^{2 x}(2 \sin x-\cos x)$

Remark: The above question can also be done as: Applying Product Rule: taking $\sin x$ as first function and $e^{2 x}$ as second function.

Put $\boldsymbol{x}=\boldsymbol{\operatorname { t a n }} \theta$. Differentiating both sides $d x=\sec ^2 \theta d \theta$.

Applying product rule

Choose the correct option in Exercises 23 and 24.

Let $\mathrm{I}=\int x^2 e^{x^3} d x=\frac{1}{3} \int e^{\left(x^3\right)}\left(3 x^2\right) d x \quad\left[\because \frac{d}{d x} x^3=3 x^2\right]$

Put $x^3=t$. Therefore $3 x^2=\frac{d t}{d x}$. Therefore $3 x^2 d x=d t$

Therefore, from (i), I $=\frac{1}{3} \int e^t d t=\frac{1}{3} e^t+\mathrm{C}$

Putting $t=x^3,=\frac{1}{3} e^{x^3}+\mathrm{C}$

Hence, Option (B) is the correct answer.

Let $\mathrm{I}=\int e^x \sec x(1+\tan x) d x=\int e^x(\sec x+\sec x \tan x) d x$ It is of the form $\int e^x\left(f(x)+f^{\prime}(x)\right) d x$

Here $f(x)=\sec x$ and so $f^{\prime}(x)=\sec x \tan x$

Hence, Option (B) is the correct answer.

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$\int \sqrt{4-x^2} d x=\int \sqrt{2^2-x^2} d x$

$\int \sqrt{1-4 x^2} d x=\int \sqrt{1^2-(2 x)^2} d x$

$\int \sqrt{x^2+4 x+6} d x$

Coefficient of $x^2$ is unity. So let us complete squares by adding and subtracting $\left(\frac{1}{2} \text { Coefficient of } x\right)^2=2^2$

$\int \sqrt{x^2+4 x+1} d x=\int \sqrt{x^2+4 x+2^2+1-4} d x$

$\int \sqrt{1-4 x-x^2} d x=\int \sqrt{-x^2-4 x+1} d x$

Making coefficient of $x^2$ unity

(Note: You can’t take this (-) sign out of this bracket because square root of -1 is imaginary)

$\int \sqrt{x^2+4 x-5} d x=\int \sqrt{x^2+4 x+2^2-4-5} d x$

$\int \sqrt{1+3 x-x^2} d x=\int \sqrt{-x^2+3 x+1} d x$

$\int \sqrt{x^2+3 x} d x=\int \sqrt{x^2+3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2} d x=\int \sqrt{\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2} d x$

$\int \sqrt{1+\frac{x^2}{9}} d x=\int \sqrt{\frac{9+x^2}{9}} d x=\int \frac{\sqrt{x^2+3^2}}{3} d x=\frac{1}{3} \int \sqrt{x^2+3^2} d x$

Choose the correct option in Exercises 10 to 11:

$\int \sqrt{1+x^2} d x=\int \sqrt{x^2+1^2} d x$

$\int_{-1}^1(x+1) d x=\left(\frac{x^2}{2}+x\right)_{-1}^1=\phi(b)-\phi(a)$

(By Second Fundamental Theorem given in Equation (i) page 496)

Remark: [Constant $c$ will never occur in the value of a definite integral because $c$ in the value of $\phi(b)$ gets cancelled with $c$ in $\phi(a)$ when we subtract them to get $\phi(b)-\phi(a)]$.

$\int_2^3 \frac{1}{x} d x=(\log |x|)_2^3=\phi(b)-\phi(a)=\log |3|-\log |2|$

$\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x=\left(4 \frac{x^4}{4}-5 \frac{x^3}{3}+6 \frac{x^2}{2}+9 x\right)_1^2$

$\int_0^{\frac{\pi}{4}} \sin 2 x d x=\left(\frac{-\cos 2 x}{2}\right)_0^{\frac{\pi}{4}}=\frac{-\cos \frac{\pi}{2}}{2}-\left(\frac{-\cos 0}{2}\right)$

$\int_0^{\frac{\pi}{2}} \cos 2 x d x=\left(\frac{\sin 2 x}{2}\right)_0^{\frac{\pi}{2}}=\frac{\sin \pi}{2}-\frac{\sin 0}{2}$

$\int_4^5 e^x d x=\left(e^x\right)_4^5=e^5-e^4=e^4(e-1)$.

$\int_0^{\frac{\pi}{4}} \tan x d x=(\log |\sec x|)_0^{\frac{\pi}{4}}$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x=(\log |\operatorname{cosec} x-\cot x|)_{\frac{\pi}{6}}^{\frac{\pi}{4}}$

$\int_0^1 \frac{d x}{\sqrt{1-x^2}}=\left(\sin ^{-1} x\right)_0^1 \quad\left[\because \int \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1} \frac{x}{a}\right]$

$\int_0^1 \frac{d x}{1+x^2}=\left(\tan ^{-1} x\right)_0^1 \quad\left[\because \int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]$

$\int_2^3 \frac{1}{x^2-1} d x=\int_2^3 \frac{1}{x^2-1^2} d x$

Evaluate the definite integrals in Exercises 12 to 20:

$\int_0^{\frac{\pi}{2}} \cos ^2 x d x=\int_0^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x=\int_0^{\frac{\pi}{2}} \frac{1}{2}(1+\cos 2 x) d x$

$\int_2^3 \frac{x}{x^2+1} d x=\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1} d x$

$=\frac{1}{2}\left(\log \left|x^2+1\right|\right)_2^3$.

(Here $f(x)=x^2+1$ and $f^{\prime}(x)=2 x$ )

$\int_0^1 \frac{2 x+3}{5 x^2+1} d x=\int_0^1\left(\frac{2 x}{5 x^2+1}+\frac{3}{5 x^2+1}\right) d x$

We now evaluate $\int_0^1 x e^{x^2} d x$

Let us first evaluate $\int x e^{x^2} d x$

Put $x^2=t$. Therefore $2 x=\frac{d t}{d x} \quad \therefore 2 x d x=d t$

Therefore, from (i), $\int x e^{x^2} d x=\frac{1}{2} \int e^t d t=\frac{1}{2} e^t$

Putting $\boldsymbol{t}=\boldsymbol{x}^{\mathbf{2}},=\frac{1}{2} e^{x^2}$

∴ The given integral $\int_0^1 x e^{x^2} d x=\frac{1}{2}\left(e^{x^2}\right)_0^1$

[By (ii)]

Note: Please note that limits 0 and 1 specified in the given integral are limits for $x$.

Therefore after substituting $x^2=t$ and evaluating the integral, we must put back $t=x^2$ and only then use $\int_a^b f(x) d x=\phi(b)-\phi(a)$.

Remark: In the next Exercise 7.10 we shall also learn to change the limits of integration from values of $x$ to values of $t$ and then we may use our discretion even here also.

The integrand $\frac{5 x^2}{(x+1)(x+3)}$ is a rational function and degree of numerator $=$ degree of denominator.

So let us apply long division.

Substituting this value back into (i),

where $I=\int_1^2 \frac{-20 x-15}{(x+1)(x+3)} d x$

Let integrand of $\mathrm{I}=\frac{-20 x-15}{(x+1)(x+3)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x+3}$

(Partial Fraction Decomposition)

Multiplying both sides by L.C.M. $=(x+1)(x+3)$,

Equating coefficients of $x$ and constant terms on both sides,

Coefficients of $\boldsymbol{x}: \mathrm{A}+\mathrm{B}=-20$

Constant terms: $3 \mathrm{~A}+\mathrm{B}=-15$

Subtracting (iv) and (v), $-2 \mathrm{~A}=-5$. Therefore $\mathrm{A}=\frac{5}{2}$.

Substituting $\mathrm{A}=\frac{5}{2}$ back into (iv), $\frac{5}{2}+\mathrm{B}=-20 \Rightarrow \mathrm{~B}=-20-\frac{5}{2}$

or $\quad \mathrm{B}=\frac{-40-5}{2}=\frac{-45}{2}$

Substituting these values of A and B back into (iii),

Substituting this value of I back into (ii),

$\int_0^{\frac{\pi}{4}}\left(2 \sec ^2 x+x^3+2\right) d x=2 \int_0^{\frac{\pi}{4}} \sec ^2 x d x+\int_0^{\frac{\pi}{4}} x^3 d x+2 \int_0^{\frac{\pi}{4}} 1 d x$

$\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x=\int_0^\pi\left[\left(\frac{1-\cos x}{2}\right)-\left(\frac{1+\cos x}{2}\right)\right] d x$

$\int_0^2 \frac{6 x+3}{x^2+4} d x=\int_0^2 \frac{6 x}{x^2+4} d x+3 \int_0^2 \frac{1}{x^2+4} d x$

$\int_0^1\left(x e^x+\sin \frac{\pi x}{4}\right) d x=\int_0^1 x e_{\text {II }}^x d x+\int_0^1 \sin \frac{\pi x}{4} d x$

Applying Product Rule on first definite integral,

For Exercises 21 and 22 — choose the correct option:

$\int_1^{\sqrt{3}} \frac{d x}{1+x^2}=\left(\tan ^{-1} x\right)_1^{\sqrt{3}}=\tan ^{-1} \sqrt{3}-\tan ^{-1} 1$

Hence, Option (D) is the correct answer.

$\int_0^{\frac{2}{3}} \frac{d x}{4+9 x^2}=\int_0^{\frac{2}{3}} \frac{d x}{(3 x)^2+2^2}=\left[\frac{1}{2} \frac{\tan ^{-1} \frac{3 x}{2}}{3 \rightarrow \text { Coefficient of } x \text { in } 3 x}\right]$

Hence, Option (C) is the correct answer.

Let $\mathrm{I}=\int_0^1 \frac{x}{x^2+1} d x=\frac{1}{2} \int_0^1 \frac{2 x}{x^2+1} d x$

Put $x^2+1=t$. Therefore $2 x=\frac{d t}{d x} \Rightarrow 2 x d x=d t$.

We convert the limits from $x$-values to $\boldsymbol{t}$-values.

When $x=0, t=0+1=1$

When $x=1, t=1+1=2$

Therefore, from (i), I $=\frac{1}{2} \int_1^2 \frac{d t}{t}=\frac{1}{2}(\log |t|)_1^2=\frac{1}{2}(\log |2|-\log |1|)$

Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi$

Put $\sin \phi=t$.

( ∵ one factor of integrand is $\cos ^5 \phi$ where $n=5$ is odd.)

Changing limits from $\phi$ to $t$:

When $\phi=0, t=\sin \phi=\sin 0=0$

When $\phi=\frac{\pi}{2}, t=\sin \phi=\sin \frac{\pi}{2}=1$

We rewrite the integrand as $\sqrt{\sin \phi} \cos ^5 \phi=\sqrt{\sin \phi} \cos ^4 \phi \cos \phi$

Let $I=\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$

Put $x=\tan \theta . \quad \therefore \quad \frac{d x}{d \theta}=\sec ^2 \theta \Rightarrow d x=\sec ^2 \theta d \theta$

Updating the limits of integration:

When $x=0, \tan \theta=0=\tan 0 \Rightarrow \theta=0$

When $x=1, \tan \theta=1=\tan \frac{\pi}{4} \Rightarrow \theta=\frac{\pi}{4}$

Using integration by parts

Let $\mathrm{I}=\int_0^2 x \sqrt{x+2} d x$

Put $\sqrt{\text { Linear }}=t$, i.e., $\sqrt{x+2}=t$. Therefore $x+2=t^2$.

Updating the limits of integration:

When $x=0, t=\sqrt{x+2}=\sqrt{2}$

When $x=2, t=\sqrt{x+2}=\sqrt{2+2}=\sqrt{4}=2$.

Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} d x=-\int_0^{\frac{\pi}{2}} \frac{-\sin x}{1+\cos ^2 x} d x$

Put $\cos x=t$. Therefore $-\sin x=\frac{d t}{d x} \Rightarrow-\sin x d x=d t$.

Updating the limits of integration:.

When $x=0, t=\cos 0=1$, When $x=\frac{\pi}{2}, t=\cos \frac{\pi}{2}=0$

$\int_0^2 \frac{d x}{4+x-x^2}=\int_0^2 \frac{d x}{-x^2+x+4}=\int_0^2 \frac{d x}{-\left(x^2-x-4\right)}$

(Making coeff. of $x^2$ numerically unity)

Completing the square by adding and subtracting

In a similar way, $(\sqrt{\mathbf{1 7}}-3)(\sqrt{\mathbf{1 7}}-1)=20-4 \sqrt{\mathbf{1 7}})$

Let $\mathrm{I}=\int_{-1}^1 \frac{d x}{x^2+2 x+5}=\int_{-1}^1 \frac{d x}{x^2+2 x+1+4}$ (completing the square)

Put $x+1=t . \quad \therefore \quad \frac{d x}{d t}=1 \quad \Rightarrow d x=d t$

Updating the limits of integration:

When $x=-1, t=-1+1=0$

When $x=1, t=1+1=2$

Let $\mathrm{I}=\int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x$

[Type $\int(f(x)+g(x)) e^{a x} d x$. Put $a x=t$ and it will become

Put $2 x=t \quad \therefore \quad 2=\frac{d t}{d x} \quad \Rightarrow \quad 2 d x=d t \quad \Rightarrow \quad d x=\frac{d t}{2}$

Updating the limits of integration:

When $x=1, t=2 x=2$, When $x=2, t=2 x=4$

Choose the correct option in Exercises 9 and 10.

Let $\mathrm{I}=\int_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{1 / 3}}{x^4} d x$

Put $\quad x^{-2}-1=t$

Therefore $-2 x^{-3}=\frac{d t}{d x} \quad \Rightarrow-2 x^{-3} d x=d t$

Updating the limits of integration:

When $x=\frac{1}{3}, t=x^{-2}-1=\left(\frac{1}{3}\right)^{-2}-1$

When $x=1, t=1^{-2}-1=1-1=0$

Hence, Option (A) is the correct answer.

$f(x)=\int_0^x t \sin t d t$

I II

Using integration by parts

Hence, Option (B) is the correct answer.

OR

[ ∵ Derivative operator and integral operator cancel with each other]

Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \cos ^2 x d x$

Adding equations (i) and (ii),

Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

Adding equations (i) and (ii), we get

Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{3 / 2} x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x} d x$

Changing $x$ to $\frac{\pi}{2}-x$

Adding equations (i) and (ii),

Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\cos ^5 x}{\sin ^5 x+\cos ^5 x} d x$

Adding equations (i) and (ii), we get

Let $\mathrm{I}=\int_{-5}^5|x+2| d x$

We first remove the absolute value sign to evaluate this integral.

Putting expression within modulus equal to 0 , we have

We know by definition of modulus function, that

[By (ii)]

[By (i)]

By using the properties of definite integrals, evaluate the integrals in Exercises 7 to 11:

Let $\mathrm{I}=\int_0^1 x(1-x)^n d x$

Let $\mathrm{I}=\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$

Changing $x$ to $\frac{\pi}{4}-x \quad\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right]$

Adding equations (i) and (ii), we get

or $\quad 2 \mathrm{I}=(\log 2)[x]_0^{\frac{\pi}{4}}=\frac{\pi}{4} \log 2 \quad$ Dividing by $2, \mathrm{I}=\frac{\pi}{8} \log 2$.

Let I $=\int_0^2 x \sqrt{2-x} d x$

Changing $x$ to $2-x$

Let $\mathrm{I}=\int_0^{\pi / 2}(2 \log \sin x-\log \sin 2 x) d x$

Adding equations (i) and (ii),

Let $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x d x \quad$ or $\quad I=2 \int_0^{\frac{\pi}{2}} \sin ^2 x d x$

$\left[\because\right.$ For $f(x)=\sin ^2 x, f(-x)=\sin ^2(-x)=(-\sin x)^2=\sin ^2 x=f(x)$

$\therefore f(x)$ is an even function of $x$ and hence

Adding equations (i) and (ii), we get

Using properties of definite integrals, evaluate the following integrals in Exercises 12 to 18:

Let $\mathrm{I}=\int_0^\pi \frac{x}{1+\sin x} d x$

Changing $x$ to $\pi-x, \mathrm{I}=\int_0^\pi \frac{\pi-x}{1+\sin (\pi-x)} d x$

or $\mathrm{I}=\int_0^\pi \frac{\pi-x}{1+\sin x} d x$

…(ii) $\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right]$

Adding equations (i) and (ii), we get

Let $\mathrm{I}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^7 x d x$

Here Integrand $f(x)=\sin ^7 x$

$\therefore f(-x)=\sin ^7(-x)=(-\sin x)^7=-\sin ^7 x=-f(x)$

$\therefore f(x)$ is an odd function of $x$.

$\int_0^{2 \pi} \cos ^5 x d x=2 \int_0^\pi \cos ^5 x d x$

Here $f(x)=\cos ^5 x \quad \therefore \quad f(2 \pi-x)=\cos ^5(2 \pi-x)=\cos ^5 x$

$\left[\because \int_0^{2 a} f(x) d x=0\right.$, if $f(2 a-x)=-f(x)$. Here $f(x)=\cos ^5 x$

$\left.\therefore f(\pi-x)=\cos ^5(\pi-x)=(-\cos x)^5=-\cos ^5 x=-f(x)\right]$

Alternatively. We now evaluate $\int_0^{2 \pi} \cos ^5 x d x$, put $\sin x=t$.

Remark: In fact $\int_0^{2 \pi} \cos ^n x d x$ or $\int_0^\pi \cos ^n x d x$ for all positive odd integers $n$ is equal to zero.

This is a very important result for I.I.T. Entrance Examination.

Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$

Replacing $x$ with $\frac{\pi}{2}-x$ in integrand of (i),

Adding equations (i) and (ii), we have

Let $\mathrm{I}=\int_0^\pi \log (1+\cos x) d x$

Adding equations (i) and (ii), we get

Dividing by $2, \mathrm{I}=\int_0^\pi \log \sin x d x=2 \int_0^{\frac{\pi}{2}} \log \sin x d x$

Adding equations (iii) and (iv), we get

Dividing by $2, \mathrm{I}=\int_0^{\frac{\pi}{2}}(\log \sin x \cos x) d x$

or $\quad \mathrm{I}=\int_0^{\frac{\pi}{2}} \log \sin 2 x d x-\frac{\pi}{2} \log 2$

or $\quad \mathrm{I}=\mathrm{I}_1-\frac{\pi}{2} \log 2$

where $\mathrm{I}_1=\int_0^{\frac{\pi}{2}} \log \sin 2 x d x$

Put $2 \boldsymbol{x}=\boldsymbol{t}$ to make $\mathrm{I}_1$ look as I given by (iii)

$\therefore \quad 2=\frac{d t}{d x} \quad$ or $\quad 2 d x=d t \quad$ or $\quad d x=\frac{d t}{2}$

To change the limits: When $x=0, \quad t=2 x=0$

When $x=\frac{\pi}{2}, t=2 x=\pi$

∴ From (vi), $\mathrm{I}_1=\int_0^\pi \log \sin t \frac{d t}{2}=\frac{1}{2} \int_0^\pi \log \sin t d t$

or

(For reason see Explanation within brackets below Equation (iii))

or $\mathrm{I}_1=\int_0^{\frac{\pi}{2}} \log \sin t d t=\int_0^{\frac{\pi}{2}} \log \sin x d x\left[\because \int_a^b f(t) d t=\int_a^b f(x) d x\right]$

or $\quad \mathrm{I}_1=\frac{\mathrm{I}}{2}$

[By Equation (iii)]

Putting this value of $\mathrm{I}_1$ in Equation (v), $\mathrm{I}=\frac{\mathrm{I}}{2}-\frac{\pi}{2} \log 2$

Multiplying both sides by the L.C.M. $=2,2 \mathrm{I}=\mathrm{I}-\pi \log 2$

or $\quad 2 \mathrm{I}-\mathrm{I}=-\pi \log 2$ or $\mathrm{I}=-\pi \log 2$.

Let $\mathrm{I}=\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$

Adding equations (i) and (ii), we get

or $2 \mathrm{I}=\int_0^a 1 d x=(x)_0^a=a \therefore \quad \mathrm{I}=\frac{a}{2}$.

Let I $=\int_0^4|x-1| d x$

Putting the expression ( $x-1$ ) within modulus equal to zero, we have

Given: $f(x)=f(a-x)$

and $g(x)+g(a-x)=4$

Let $\mathrm{I}=\int_0^a f(x) g(x) d x$

$\therefore \quad \mathrm{I}=\int_0^a f(a-x) \mathrm{g}(a-x) d x \quad\left[\because \int_0^a \mathrm{~F}(x) d x=\int_0^a \mathrm{~F}(a-x) d x\right]$

Putting $f(a-x)=f(x)$ from ( $i$ ),

Adding equations (iii) and (iv), we get

$2 \mathrm{I}=\int_0^a(f(x) g(x)+f(x) g(a-x)) d x=\int_0^a f(x)(g(x)+g(a-x)) d x$

or 2I $=\int_0^a f(x)(4) d x \quad$ [By (ii)] $=4 \int_0^a f(x) d x$

Dividing by $2, \mathrm{I}=2 \int_0^a f(x) d x=$ R.H.S.

For Exercises 20 and 21 — choose the correct option:

Let $\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^3+x \cos x+\tan ^5 x+1\right) d x$

∵ Each of the three functions $x^3, x \cos x$ and $\tan ^5 x$ is an odd function of $x$ as $f(-x)=-f(x)$ for each of them and $\int_{-a}^a f(x) d x=0$ for each odd function $\left.f(x)\right]$

Hence, Option (C) is the correct option.

Let $I=\int_0^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$

or $\quad \mathrm{I}=\int_0^{\frac{\pi}{2}} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x$

Adding equations (i) and (ii), we get

The integrand $\frac{1}{x-x^3}$ is a rational function of $x$ and the

denominator $x-x^3=x\left(1-x^2\right)=x(1-x)(1+x)$ is the product of more than one factor. So, will form partial fractions.

Multiplying every term of Equation (i) by L.C.M. $\quad=x(1-x)(1+x)$,

Equating coefficients of like powers on both sides,

Constants: $\mathrm{A}=1$

Substituting $\mathrm{A}=1$ back into (ii), $-1+\mathrm{B}-\mathrm{C}=0$ or $\mathrm{B}-\mathrm{C}=1 \ldots(i v)$

Adding equations (iii) and (iv), $2 \mathrm{~B}=1 \Rightarrow \mathrm{~B}=\frac{1}{2}$

From (iii), $\mathrm{C}=-\mathrm{B}=\frac{-1}{2}$

Substituting these values of A, B, C back into (i),

$\int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} d x$

$\mathrm{I}=\int \frac{d x}{x \sqrt{a x-x^2}} \quad\left[\right.$ Form $\int \frac{\boldsymbol{d} \boldsymbol{x}}{\text { Linear } \sqrt{\text { Quadratic }}}$

Put Linear $=\frac{1}{t}$, i.e., $x=\frac{1}{t}=t^{-1}$.

Differentiating both sides $d x=-\frac{1}{t^2} d t$

$\mathrm{I}=\int \frac{d x}{x^2\left(x^4+1\right)^{3 / 4}}=\int \frac{d x}{x^2\left[x^4\left(1+\frac{1}{x^4}\right)\right]^{3 / 4}}=\int \frac{d x}{x^2 \cdot x^3\left(1+\frac{1}{x^4}\right)^{3 / 4}}$

Put $1+\frac{1}{x^4}=t \quad$ or $\quad 1+x^{-4}=t$.

Differentiating both sides, $-4 x^{-5} d x=d t$

or $-\frac{4}{x^5} d x=d t \quad$ or $\quad \frac{1}{x^5} d x=-\frac{1}{4} d t$

$\therefore \quad \mathrm{I}=-\frac{1}{4} \int t^{-3 / 4} d t=-\frac{1}{4} \cdot \frac{t^{1 / 4}}{1 / 4}+c=-\left(1+\frac{1}{x^4}\right)^{1 / 4}+c$.

Here the denominators of fractional powers $\frac{1}{2}$ and $\frac{1}{3}$ of $x$ are 2 and 3. L.C.M. of 2 and 3 is 6 .

Put $\boldsymbol{x}=\boldsymbol{t}^6$. Differentiating both sides, $d x=6 t^5 d t$

Putting $t=x^{1 / 6} \quad\left(\because x=t^6 \Rightarrow t=x^{1 / 6}\right)$

Let $\mathrm{I}=\int \frac{5 x}{(x+1)\left(x^2+9\right)} d x$

Let $\frac{5 x}{(x+1)\left(x^2+9\right)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+9}$

L.C.M. $=(x+1)\left(x^2+9\right)$

Multiplying every term of (ii) by L.C.M.,

or $5 x=\mathrm{A} x^2+9 \mathrm{~A}+\mathrm{B} x^2+\mathrm{B} x+\mathrm{C} x+\mathrm{C}$

Equating coefficients of $x^2, x$ and constant terms on both sides,

$\boldsymbol{x}^{\mathbf{2}}$ :

$x$ :

Constant terms : 9A + C = 0

We now solve Equations (iii), (iv) and (v) to determine A, B, C. (iii) – (iv) gives, (to eliminate B ), $\mathrm{A}-\mathrm{C}=-5$

Adding (v) and (vi),

Substituting $\mathrm{A}=\frac{-1}{2}$ back into (iii), $\frac{-1}{2}+\mathrm{B}=0 \Rightarrow \mathrm{~B}=\frac{1}{2}$

Substituting $\mathrm{B}=\frac{1}{2}$ back into (iv), $\frac{1}{2}+\mathrm{C}=5 \Rightarrow \mathrm{C}=5-\frac{1}{2}=\frac{9}{2}$

Substituting these values of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ back into (ii),

$\int \frac{\sin x}{\sin (x-a)} d x=\int \frac{\sin (x-a+a)}{\sin (x-a)} d x$

$\int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x=\int \frac{e^{\log x^5}-e^{\log x^4}}{e^{\log x^3}-e^{\log x^2}} d x \quad\left[\because \quad n \log m=\log m^n\right]$

Let $\mathrm{I}=\int \frac{\cos x}{\sqrt{4-\sin ^2 x}} d x$

Put $\sin \boldsymbol{x}=\boldsymbol{t}$. Therefore $\cos x=\frac{d t}{d x} \Rightarrow \cos x d x=d t$

Let $\mathrm{I}=\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$

Now numerator of integrand $=\sin ^8 x-\cos ^8 x$

$=1\left[-\left(\cos ^2 x-\sin ^2 x\right)\right]\left(1-2 \sin ^2 x \cos ^2 x\right)$

$\Rightarrow \sin ^8 x-\cos ^8 x=-\cos 2 x\left(1-2 \sin ^2 x \cos ^2 x\right)$

Putting this value of $\sin ^8 x-\cos ^8 x$ in numerator of (i),

Let $\mathrm{I}=\int \frac{1}{\cos (x+a) \cos (x+b)} d x$

We know that $(x+a)-(x+b)=x+a-x-b=a-b$

Dividing and multiplying by $\boldsymbol{\operatorname { s i n }}(\boldsymbol{a}-\boldsymbol{b})$ in (i),

Integrate the functions in Exercises 12 to 22:

Let $\mathrm{I}=\int \frac{x^3}{\sqrt{1-x^8}} d x=\frac{1}{4} \int \frac{4 x^3}{\sqrt{1-\left(x^4\right)^2}} d x$

Put $\boldsymbol{x}^4=\boldsymbol{t}$. Therefore $4 x^3=\frac{d t}{d x} \Rightarrow 4 x^3 d x=d t$

Therefore, from (i), I $=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^2}}=\frac{1}{4} \sin ^{-1} t+c$

or

Let $\mathrm{I}=\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)} d x$

[Rule for evaluating $\int f\left(e^x\right) d x$, put $e^x=t$ ]

Put $e^x=t$. Therefore $e^x=\frac{d t}{d x} \quad \Rightarrow e^x d x=d t$

Now $t+2-(t+1)=t+2-t-1=1$

Replacing 1 in the numerator of integrand in (ii) by (this)

Putting $t=e^x,=\log \left|\frac{e^x+1}{e^x+2}\right|+c \quad=\log \left(\frac{e^x+1}{e^x+2}\right)+c$. $\left[\because \quad e^x+1>0\right.$ and $e^x+2>0$ and $|t|=t$ if $t \geq 0$ ]

Let $\mathrm{I}=\int \frac{1}{\left(x^2+1\right)\left(x^2+4\right)} d x$

Put $x^2=\mathrm{y}$ only in the integrand.

Now the integrand is $\frac{1}{(y+1)(y+4)}$

Let $\frac{1}{(y+1)(y+4)}=\frac{\mathrm{A}}{y+1}+\frac{\mathrm{B}}{y+4}$

Multiplying both sides by the L.C.M. $=(y+1)(y+4)$,

$1=\mathrm{A}(y+4)+\mathrm{B}(y+1)$

or $1=\mathrm{A} y+4 \mathrm{~A}+\mathrm{B} y+\mathrm{B}$

comparing coefficient of $\mathrm{y}, \mathrm{A}+\mathrm{B}=0$

comparing constants, $4 \mathrm{~A}+\mathrm{B}=1$

We now solve (iii) and (iv) to determine A and B .

(iv) – (iii) gives $3 \mathrm{~A}=1 \quad \therefore \mathrm{~A}=\frac{1}{3}$

From (iii) $\mathrm{B}=-\mathrm{A}=-\frac{1}{3}$

Substituting values of $\mathrm{A}, \mathrm{B}$ and y back into (ii),

Substituting this value back into (i),

Let $\mathrm{I}=\int \cos ^3 x e^{\log \sin x} d x \quad=\int \cos ^3 x \sin x d x$

Put $\cos x=t . \quad \therefore \quad-\sin x=\frac{d t}{d x} \quad \Rightarrow \quad-\sin x d x=d t$

$\therefore$ From $(i), \mathrm{I}=-\int t^3 d t=\frac{-t^4}{4}+c=\frac{-1}{4} \cos ^4 x+c$.

Let $\mathrm{I}=\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=\int \frac{e^{\log x^3}}{x^4+1} d x \quad=\int \frac{x^3}{x^4+1} d x$

Put $x^4+1=t$. Therefore $4 x^3=\frac{d t}{d x} \Rightarrow 4 x^3 d x=d t$

Therefore, from (i), I $=\frac{1}{4} \int \frac{d t}{t}=\frac{1}{4} \log |t|+c$

Putting $t=x^4+1,=\frac{1}{4} \log \left|x^4+1\right|+c \quad=\frac{1}{4} \log \left(x^4+1\right)+c$.

Let $\mathrm{I}=\int f^{\prime}(a x+b)(f(a x+b))^n d x$

Put $f(a x+b)=t$. Therefore $f^{\prime}(a x+b) \frac{d}{d x}(a x+b)=\frac{d t}{d x}$

$\Rightarrow a f^{\prime}(a x+b) d x=d t$

Therefore, from (i), I $=\frac{1}{a} \int t^n d t=\frac{1}{a} \frac{t^{n+1}}{n+1}+c$ if $n \neq-1$

$I=\int \frac{d x}{\sqrt{\sin ^3 x \sin (x+\alpha)}}=\int \frac{d x}{\sqrt{\sin ^3 x(\sin x \cos \alpha+\cos x \sin \alpha)}}$

Put $\cos \alpha+\cot x \sin \alpha=t$. Differentiating both sides