Indefinite Integration

$ \begin{aligned} &\\ &\begin{aligned} & \ \text { Let } I=\int \frac{2 \cos 2 x}{(1+\sin 2 x)(1+\cos 2 x)} d x \\ & =\int \frac{2 \times\left(2 \cos ^2 x-1\right) d x}{(\sin x+\cos x)^2 \times 2 \cos ^2 x} \\ & =\int \frac{\left(2 \cos ^2 x-1\right)}{(\sin x+\cos x)^2} \sec ^2 x d x \\ & =\int \frac{\left(2-\sec ^2 x\right)}{(1+\tan x)^2} \sec ^2 x d x \\ & =\int \frac{\left(1-\tan ^2 x\right)}{(1+\tan x)^2} \sec ^2 x d x \\ & =\int\left(\frac{1-\tan x}{1+\tan x}\right) \sec ^2 x d x \\ & =\int\left(\frac{1-t}{1+t}\right) d t[\because \tan x=1]\\ & =-\int\left(\frac{t-1}{1+t}\right) d t=-\int\left(1-\frac{2}{1+t}\right) d t \\ & =-t+2 \log (1+t)+C \\ & =-\tan x+2 \log (1+\tan x)+C \end{aligned} \end{aligned} $

$ \begin{aligned} & \text {Let } I=\int\left(\frac{x}{x \cos x-\sin x}\right)^2 d x \\ & =\int \frac{x^2}{(x \cos x-\sin x)^2} d x \\ & =\int(x \operatorname{cosec} x) \cdot \frac{(x \sin x)}{(x \cos x-\sin x)^2}...(i) \end{aligned} $

Let, $I_1=\int \frac{x \sin x}{(x \cos x-\sin x)^2} d x$

Let, $x \cos x-\sin x=t \Rightarrow x \sin x d x=-d$

$ \begin{aligned} \Rightarrow \quad & =-\int \frac{1}{t^2} d t \\ & =\frac{1}{x \cos x-\sin x} \end{aligned} $

From Eq. (i), Integrate $I$, we get

$\begin{aligned} & I=(x \operatorname{cosec} x) \int \frac{x \sin x d x}{(x \cos x-\sin x)^2} \\ & \quad-\int\left[(x \operatorname{cosec} x)^2 \int \frac{x \sin x d x}{(x \cos x-\sin x)^2}\right] d x \\ &= \frac{x \operatorname{cosec} x}{x \cos x-\sin x} \\ & \quad-\int\left[\frac{-x \operatorname{cosec} x \cdot \cot x+\operatorname{cosec} x}{x \cos x-\sin x}\right] d x \\ &= \frac{x \operatorname{cosec} x}{x \cos x-\sin x}+\int \frac{1}{\sin ^2 x} d x+C \\ &= \frac{x \operatorname{cosec} x}{x \cos x-\sin x}-\cot x+C\end{aligned}$

We have,

$ \begin{aligned} & \operatorname{Iet} I=\int \frac{1}{x^3\left(x^3+1\right)^{1/5}} d x \\ & =\int \frac{1}{x^2\left(1+\frac{1}{x^3}\right)^{1 / 5}} d x \end{aligned} $

Let $1+\frac{1}{x^3}=t^3$

$\begin{aligned}-\frac{5}{x^6} d x & =5 t^4 d t \Rightarrow \frac{1}{x^6} d x=-t^4 d t \\ I & =-\int \frac{1}{t} t^4 d t \\ & =-\int t^3 d t=-\frac{t^4}{4}+C \\ & =-\frac{1}{4}\left[1+\frac{1}{x^5}\right]^{4 / 5}+C \\ & =-\frac{1}{4} \frac{\left(x^5+1\right)^{4 / 5}}{x^4}+C\end{aligned}$

$\operatorname{Let} t=\int \frac{x+1}{\sqrt{x^2+x+1}} d x$

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

$I=I_1+I_2 (Let) \quad \ldots \text { (i) } $

Let $I_1=\frac{1}{2} \int \frac{2 x+1}{\sqrt{x^2+x+1}} d x$

$\operatorname{Let} x^2+x+1=t^2$

$ \begin{aligned} (2 x+1) d r & =2 d t \\ t_1 & =\frac{1}{2} \int^{\frac{1}{t}} \frac{x d t}{} \\ & =\int d t=t=\sqrt{x^2+x+1} \\ I_2 & =\frac{1}{2} \int \frac{1}{\sqrt{x^2+x+1}} d r \\ & =\frac{1}{2} \int \frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}} d x \end{aligned} $

Let $x+\frac{1}{2}=\cos d x=d x$

$ \begin{aligned} f_2 & =\frac{1}{2} \int \frac{1}{\sqrt{N^2+\frac{3}{4}}} \\ & =\frac{1}{2} \sinh ^{-1}\left(\frac{\pi}{\sqrt{3}}\right) \\ & =\frac{1}{2} \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right) \end{aligned} $

$ \begin{aligned} & \text { Hence. } \\ & I=\sqrt{x^2+x+1}+\frac{1}{2} \sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \end{aligned} $

$ \begin{aligned} \operatorname{Let} l & =\int\left(\tan ^3 x+\tan x\right) d x \\ & =\int \tan x\left(\tan ^6 x+1\right) d x \\ & =\int \tan x\left(\left(\tan ^2 x\right)^3+1^3\right) d x \\ & =\int \tan ^x\left(\tan ^2 x+1\right)\left(\tan ^4 x+1-\tan ^2 x\right) d x \\ & =\int\left(\tan ^3 x-\tan ^3 x+\tan x\right) \sec ^3 x d x \end{aligned} $

$\begin{aligned} & \text { Let } \tan x=t \\ & \sec ^2 x, d x=d t \\ & I=\int\left(t^5-t^3+t\right) d t \\ & =\frac{t^6}{6}-\frac{t^4}{4}+\frac{t^2}{2}+C=\frac{t^2\left(2 t^3-3 t^2+6\right)}{12}+C \\ & =\frac{\tan ^2 x}{12}\left(2 \tan ^3 x-3 \tan ^2 x+6\right)+C\end{aligned}$

Let

$ \begin{aligned} I & =\int \frac{\operatorname{cosec} x}{3 \cos x+4 \sin x} d x \\ & =\int \frac{\operatorname{cosec}^2 x}{3 \cot x+4} d x \end{aligned} $

$ \begin{aligned} & \text { Let } 3 \cot x+4=t \\ & \begin{aligned} &-3 \operatorname{cosec}^2 x d x=d t \\ & \operatorname{cosec}^2 x d x=\frac{-1}{3} d t \\ & I=-\frac{1}{3} \int \frac{1}{t} d t \\ &=\frac{-1}{3} \log |t|+C=\frac{1}{3} \log \left|\frac{1}{t}\right|+C \\ &=\frac{1}{3} \log \left|\frac{1}{3 \cot x+4}\right|+C \\ &=\frac{1}{3} \log \left|\frac{\sin x}{3 \cos x+4 \sin x}\right|+C \end{aligned} \end{aligned} $

Let $I=\int e^{2 x+3} \sin 6 x d x$

$ \begin{aligned} & =\frac{e^{2 x+3} \sin 6 x}{2}-\int 3 e^{2 x+3} \cdot \cos 6 x d x \\ & =\frac{e^{2 x+3} \sin 6 x}{2} \\ & -\frac{\left[\frac{3 e^{2 x+3} \cos 6 x}{2}-\int-9 e^{2 x+3} \sin 6 x d x\right]}{2}=\frac{e^{2 x+3} \sin 6 x}{2}-\frac{3 e^{2 x+3} \cos 6 x}{2}-9 I \\ & \Rightarrow 10 I=\frac{e^{2 x+3}(\sin 6 x-3 \cos 6 x)}{2} \\ & \therefore \quad I=\frac{e^{2 x+3}(\sin 6 x-3 \cos 6 x)}{20}+C \end{aligned} $

Here, $\frac{2 x^2+1}{x^3-1}=\frac{A}{x-1}+\frac{B x+C}{x^2+x+1}$

Resolving into partial fractions,

$\frac{2 x^2+1}{x^3-1}=\frac{A\left(x^2+x+1\right)+(B x+C)(x-1)}{(x-1)\left(x^2+x+1\right)}$

$\begin{aligned} & \Rightarrow \quad 2 x^2+1=A x^2+A x+A+B x^2-B x+C x-C \\ & \Rightarrow \quad 2 x^2+1=x^2(A+B)+x(A-B+C)+A-C \end{aligned}$

Comparing with coefficient, we get

$A+B=2, A-B+C=0 \text { and } A-C=1$

Clearly, $A=1, B=1$ and $C=0$

So, $7 A+2 B+C=7(l)+2(l)+0=9$

$\begin{aligned} & \int \frac{3 x+4}{x^3-2 x+4} d x=\log f(x)+C \\ & I=\int \frac{3 x+4}{x^3-2 x+4} \end{aligned}$

$\begin{aligned} & \text { Since, } x^3-2 x-4=(x-2)\left(x^2+2 x+2\right) \\ & \frac{3 x+4}{x^3-2 x+4}=\frac{A}{(x-2)}+\frac{B x+C}{\left(x^2+2 x+2\right)} \\ & \Rightarrow 3 x+4=A\left(x^2+2 x+2\right)+(B x+C)(x-2) \end{aligned}$

On putting $x=2, A=1$ and comparing coefficients of $x^2$, we get

$0=A+B \Rightarrow B=-1$

On putting $x=0$, we have $2 A-2 C$

$\Rightarrow \quad C=-1$

$\begin{array}{r} \int \frac{3 x+4}{x^3-2 x+4} d x=\int \frac{d x}{x-2}-\int \frac{x+1}{x^2+2 x+2} d x \\ =\log |x-2|-\frac{1}{2} \log \left|\left(x^2+2 x+2\right)\right|+C \\ =\log f(x)+C=\log \frac{|x-2|}{\sqrt{\left(x^2+2 x+2\right)}}+C \end{array}$

$\text { So, } \begin{aligned} f(x) & =\frac{|x-2|}{\sqrt{\left(x^2+2 x+2\right)}} \\ f(3) & =\frac{|3-2|}{\sqrt{\left((3)^2+2(3)+2\right)}} \\ f(3) & =\frac{1}{\sqrt{9+6+2}}=\frac{1}{\sqrt{17}} \end{aligned}$

Let

$I=\int \frac{e^{\tan ^{-1}(x)}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$

On putting $\tan ^{-1} x=t$ and $\frac{d x}{1+x^2}=d t$, we get

$I=\int e^t\left[t^2+2 t\right] d t \Rightarrow I=\int\left(t^2 e^t+2 e^t t\right) d t$

By integration by parts, we get

$\begin{aligned} & I=t^2 e^t-\int 2 t e^t d t-\int 2 t e^t d t+C \\ & I=t^2 e^t+C=e^{\operatorname{tin}^{-1}(x)}\left(\tan ^{-1} x\right)^2+C \end{aligned}$

$\begin{aligned} \text { Let } I & =\int \frac{d x}{(x+1)^{\frac{6}{5}}(x-3)^{\frac{4}{5}}} \\ I & =\int \frac{d x}{(x+1)^2\left(\frac{x-3}{x+1}\right)^{\frac{4}{5}}} \end{aligned}$

On putting $\frac{(x-3)}{(x+1)}=t$, then

$\begin{aligned} & d t=\frac{4}{(x+1)^2} d x \\ & I=\int \frac{d t}{4 t^{\frac{4}{5}}}=\frac{5}{4} t^{\frac{1}{5}}+C=\frac{5}{4}\left(\frac{x-3}{x+1}\right)^{\frac{1}{5}}+C \end{aligned}$

Here, $I_n=\int\left(\cos ^n x+\sin ^n x\right) d x$

$I_n=\int \cos ^{n-1} x \cos x d x+\int \sin ^{n-1} x \sin x d x$

Using by parts,

$\begin{gathered} \left.I_n=\cos ^{n-1} x \int \cos x d x-\int\left(\frac{d}{d x}\left(\cos ^{n-1} x\right) \int \cos x d x\right)\right) d x \\ +\sin ^{n-1} x \int \sin x d x-\int\left(\frac{d}{d x}\left(\sin ^{n-1}\right) \int\left(\int \sin d x\right) d x\right. \\ \Rightarrow I_n=\cos ^{n-1} \sin x+\int(n-1) \cos ^{n-2} x \sin ^2 x d x \\ -\sin ^{n-1} x \cos x+\int(n-1) \sin ^{n-2} x \cos ^2 x d x \\ \Rightarrow I_n=\cos ^{n-1} x \sin x-\sin ^{n-1} x \cos x \\ +\int(n-1) \cos ^{n-2} x\left(1-\cos ^2 x\right) d x \\ +\int(n-1) \sin ^{n-2} x\left(1-\sin ^2 x\right) d x \end{gathered}$

$\begin{aligned} \Rightarrow I_n=\cos ^{n-1} x \sin x-\sin ^{n-1} x & \cos x \\ & +\int(n-1) \cos ^{n-2} x d x \\ -\int(n-1) \cos ^n x d x+ & \int(n-1) \sin ^{n-2} x d x \\ & -\int(n-1) \sin ^n x d x \end{aligned}$

$\begin{aligned} & \Rightarrow I_n=\cos ^{n-1} x \sin x-\sin ^{n-1} x \cos x \\ & +\int(n-1) \cos ^{n-2} x d x+\int(n-1) \sin ^{n-2} x d x \\ & -[(n-1) \int\left(\cos ^n x+\sin ^n x\right) d x \\ \end{aligned}$

$\begin{array}{rl} \Rightarrow I_n=\cos ^{n-1} x \sin x-\sin ^{n-1} & x \cos x \\ & +(n-1) I_{n-2}-(n-1) I_n \end{array}$

$\begin{aligned} & \Rightarrow I_n(1+n-1)-(n-1) I_{n-2} \\ &=\cos ^{n-1} x \sin x-\sin ^{n-1} x \cos x \end{aligned}$

$\begin{aligned} & \Rightarrow n\left(l_n\right)-(n-1) I_{n-2}=\cos ^{n-1} x \sin x-\sin ^{n-1} x \cos x \\ & \Rightarrow I_n-\left(\frac{n-1}{n}\right) I_{n-2}=\frac{\sin x \cos x}{n}\left(\cos ^{n-2} x-\sin ^{n-2} x\right) \\ & \quad=\frac{\sin x \cos x}{n} f(x) \\ & \therefore f(x)=\cos ^{n-2} x-\sin ^{n-2} x \end{aligned}$

$\begin{aligned} & f(x)=\int x^2 \cos ^2 x\left(2 x \tan ^2 x-2 x-6 \tan x\right) d x \\ & =\int\left(2 x^3 \sin ^2 x-2 x^3 \cos ^2 x\right) d x-3 \int x^2(2 \sin x \cdot \cos x) d x \\ & =-\int 2 x^3 \cdot \cos 2 x d x-3 \int x^2 \sin 2 x d x \\ & =-\left[2 x^3 \int \cos 2 x d x-\int 2 \cdot 3 x^2 \frac{\sin 2 x}{2} d x\right]-3 \int x^2 \sin 2 x d x \\ & =-x^3 \sin 2 x+3 \int x^2 \sin 2 x d x-3 \int x^2 \sin 2 x d x \\ & \quad f(x)=-x^3 \sin 2 x+C \\ & \therefore \quad f(0)=\pi \Rightarrow C=\pi \\ & \therefore \quad f(x)=-x^3 \sin 2 x+\pi \end{aligned}$

$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}(x+\sqrt{x}) d x$

$=\int\left(\sqrt{x} e^{\sqrt{x}}+e^{\sqrt{x}}\right) d x=\int \sqrt{x} e^{\sqrt{x}} d x+\int e^{\sqrt{x}} d x$

Now, $\int \sqrt{x} \cdot e^{\sqrt{x}} d x$

$\begin{aligned} \text { Let } x & =p^2 \\ \Rightarrow d x & =2 p d p \end{aligned}$

$\begin{aligned} & =\int \sqrt{x} e^{\sqrt{x}} d x=\int p e^p(2 p d p)=2 \int p^2 e^p d p \\ & =2\left[p^2 e^p-2 \int p e^p d p\right] \\ & =2\left[p^2 e^p-2\left\{p e^p-e^p\right\}\right]+K \\ & =2\left[p^2 p^p-2 p e^p+2 e^p\right]+K \\ & =2\left[x e^{\sqrt{x}}-2 \sqrt{x} e^{\sqrt{x}}+2 e^{\sqrt{x}}\right]+K \end{aligned}$

$\begin{aligned} & \int e^{\sqrt{x}} d x=2 \int p e^p d p=2\left[p e^p-e^p\right]=2\left[\sqrt{x} e^{\sqrt{x}}-e^{\sqrt{x}}\right] \\ & \therefore \int \frac{e^{\sqrt{x}}}{\sqrt{x}}(x+\sqrt{x}) d x \\ & =2 x e^{\sqrt{x}}-4 \sqrt{x} e^{\sqrt{x}}+4 e^{\sqrt{x}}+2 \sqrt{x} e^{\sqrt{x}}-2 e^{\sqrt{x}}+K \\ & =e^{\sqrt{x}}(2 x-2 \sqrt{x}+2)+K \\ & \therefore A=2 B=-2, C=2 \\ & \therefore A+B+C=2 \end{aligned}$

$\int \frac{1+\sqrt{\tan x}}{\sin 2 x} d x$

$\begin{aligned} & =\int \operatorname{cosec} 2 x d x+\int \frac{\sqrt{\tan x}}{\sin 2 x} d x \Rightarrow \int \frac{1+\sqrt{\tan x}}{\sin x} d x \\ & =\int \operatorname{cosec} 2 x d x+\frac{\sqrt{\tan x}}{\sin 2 x} d x \\ & =\frac{1}{2} \log |\tan x|+\frac{1}{2} \int \frac{\sqrt{\sin x}}{\sqrt{\cos x} \sin x \cos x} d x \\ & =\frac{1}{2} \log |\tan x|+\frac{1}{2} \int \frac{d x}{\sqrt{\cos x} \sqrt{\sin x}} \cos x \frac{\sqrt{\cos x}}{\sqrt{\cos x}} \\ & =\frac{1}{2} \log |\tan x|+\frac{1}{2} \int \frac{\sec ^2 x d x}{\sqrt{\tan x}} \\ & =\frac{1}{2} \log |\tan x|+\frac{1}{2}(2) \cdot \sqrt{\tan x}+C \end{aligned}$

$\begin{aligned} & =\frac{1}{2} \log |\tan x|+\sqrt{\tan x}+C \\ & \therefore 4 A-2 B=4 \times \frac{1}{2}-2(l)=2-2=0 \end{aligned}$

$\begin{aligned} & \int \frac{1+\tan x \cdot \tan (x+a)}{\tan x \cdot \tan (x+a)} d x \\ & =\int \frac{\cos (x) \cos (x+a)+\sin x \sin (x+a)}{\sin x \sin (x+a)} d x \\ & =\int \frac{\cos (a)}{\sin x \cdot \sin (x+a)} d x=\frac{\cos a}{\sin a} \int \frac{\sin (x+a-x)}{\sin x \cdot \sin (x+a)} d x \\ & =\cot a \int \frac{\sin (x+a) \cos x-\cos (x+a) \sin x}{\sin x \sin (x+a)} d x \\ & =\cot a\left[\int \cot x d x-\int \cot (x+a) d x\right]+C \\ & =\cot a[\log |\sin x|-\log |\sin (x+a)|]+C \end{aligned}$

$\begin{aligned} & I_n=\int \cot ^n x d x \\ & I_6+I_4=\int \cot ^6 x d x+\int \cot ^4 x d x \\ & =\int \cot ^4 x\left(\cot ^2 x+1\right) d x \\ & =-\int \cot ^4 x \cdot\left(-\operatorname{cosec}^2 x\right) d x \\ & =-\int(\cot x)^4 \cdot\left(-\operatorname{cosec}^2 x\right) d x=-\frac{(\cot x)^5}{5} \\ \end{aligned}$

$\therefore$ Assertion is true.

$\begin{array}{rl} \int \cot ^n & x d x=\int \cot ^{n-2} x \cdot \cot ^2 x d x \\ & =\int \cot ^{n-2} x \cdot\left(\operatorname{cosec}^2 x-1\right) d x \\ & =\int \cot ^{n-2} x \cdot \operatorname{cosec}^2 x \cdot d x-\int \cot ^{n-2} x d x \\ & =-\int(\cot x)^{n-2} x \cdot\left(-\operatorname{cosec}^2 x\right) d x-\int \cot ^{n-2} x d x \\ & =-\frac{(\cot x)^{n-1}}{n-1}-\int \cot ^{n-2} x d x \end{array}$

$\therefore$ Reason is false.

$\begin{aligned} & I_n=\int \tan ^n x d x \\ & I_0+I_1+2 I_2+2 I_3+2 I_4+I_5+I_6 \\ & =\int \tan ^0 x d x+\int \tan ^1 x d x+2 \int \tan ^2 x d x+2 \int \tan ^3 x d x \\ & \quad+2 \int \tan ^4 x d x+\int \tan ^5 x \cdot d x+\int \tan ^6 x d x \\ & \begin{array}{r} =\int\left[1+\tan x+2 \tan ^2 x+2 \tan ^3 x+2 \tan ^4 x+\tan ^5 x\right. \\ \left.\quad+\tan ^6 x\right] d x \end{array} \\ & \begin{array}{r} =\int\left[1+\tan ^2 x+\left(\tan x+\tan ^3 x\right)+\left(\tan ^2 x+\tan ^4 x\right)\right. \end{array} \\ & \left.\quad+\left(\tan ^3 x+\tan ^5 x\right)+\left(\tan ^4 x+\tan ^6 x\right)\right] d x \\ & =\int\left[1+\tan ^2 x+\tan x\left(1+\tan ^2 x\right)+\tan ^2 x\left(1+\tan ^2 x\right)\right. \\ & \left.\quad+\tan ^3 x\left(1+\tan ^2 x\right)+\tan ^4 x\left(1+\tan ^2 x\right)\right] d x \\ &=\int\left[\sec ^2 x+\tan x \sec ^2 x+\tan ^2 x \cdot \sec ^2 x\right. \\ &\left.+\tan ^3 x \sec ^2 x+\tan ^4 x \sec ^2 x\right] d x \end{aligned}$

$\begin{aligned} & =\int\left[1+\tan x+\tan ^2 x+\tan ^3 x+\tan ^4 x\right] \sec ^2 x d x \\ & =\int \sec ^2 x \cdot d x+\int\left(\tan x+\tan ^2 x+\tan ^3 x+\tan ^4 x\right) \sec^2x dx\\ & =\frac{\tan x}{1}+\frac{\tan ^2 x}{2}+\frac{\tan ^3 x}{3}+\frac{\tan ^4 x}{4}+\frac{\tan ^5 x}{5} \\ & =\sum_{k=1}^5 \frac{\tan ^k x}{k}=\sum_{k=1}^n \frac{\tan ^k x}{k} \Rightarrow n=5 \end{aligned}$

$\begin{aligned} \text {Let } I & =\int \frac{e^{\cos x}}{\sin ^2 x}(2 \log \operatorname{cosec} x+\sin 2 x) d x \\ I & =\int e^{\cos x}\left(\log \operatorname{cosec}^2 x+\sin 2 x\right) \cdot \operatorname{cosec}^2 x d x \end{aligned}$

Let $\cot x=t \Rightarrow-\operatorname{cosec}^2 x d x=d t$

$\begin{aligned} & I=-\int e^t\left[\log \left(1+t^2\right)+\frac{2 t}{1+t^2}\right] d t \\ & {\left[\text { Here, } \frac{2 t}{1+t^2}=\frac{2 \cot x}{1+\cot ^2 x}=\frac{2 \cot x}{\operatorname{cosec}^2 x}\right.} \\ & \left.=\frac{2 \cos x / \sin x}{1 / \sin ^2 x}=\sin 2 x\right] \\ \end{aligned}$

$ \begin{aligned} & \Rightarrow \quad I=-e^t \cdot \log \left(1+t^2\right)+C \\ & {\left[\int e^x\left(f(x)+f^{\prime}(x)\right) d x=e^x \cdot f(x)+C\right]} \\ & \Rightarrow \quad I=-e^{\cot x} \log \operatorname{cosec}^2 x+C \\ & \Rightarrow \quad I=-2 e^{\cot x} \log \operatorname{cosec} x+C \end{aligned}$

$\begin{aligned} & x=\frac{t^3}{t^2-1}, y=\frac{t}{t^2-1} \text { and let } I=\int \frac{d x}{x-3 y} \\ & x=\frac{t^3}{t^2-1} \end{aligned}$

On differentiating w.r.t. $t$, we get

$\begin{aligned} \frac{d x}{d t} & =\frac{\left(t^2-1\right)\left(3 t^2\right)-t^3(2 t)}{\left(t^2-1\right)^2} \\ \Rightarrow \quad \frac{d x}{d t} & =\frac{3 t^4-3 t^2-2 t^4}{\left(t^2-1\right)^2} \\ \Rightarrow \quad d x & =\frac{t^4-3 t^2}{\left(t^2-1\right)^2} d t \\ \Rightarrow \quad d x & =\frac{t^2\left(t^2-3\right)}{\left(t^2-1\right)^2} d t \end{aligned}$

$\begin{aligned} \Rightarrow \quad I & =\int \frac{1}{\frac{t^3}{t^2-1}-\frac{3 t}{t^2-1}} \cdot \frac{t^2\left(t^2-3\right)}{\left(t^2-1\right)^2} d t \\ & =\int \frac{\left(t^2-1\right)}{t\left(t^2-3\right)} \cdot \frac{t^2\left(t^2-3\right)}{\left(t^2-1\right)^2} d t \\ & =\int \frac{t}{t^2-1} d t=\frac{1}{2} \int \frac{2 t}{t^2-1} d t \\ & =\frac{1}{2} \log \left(t^2-1\right)+C \end{aligned}$

${{2{x^4} - {x^3} + 3{x^2} - x + 4} \over {{x^2} - 3x + 2}}$

$\because$ Degree of numerator is greater than the degree of denominator.

$\therefore$ First we divide,

$\begin{aligned} \text { or } & =\frac{2 x^4-x^3+3 x^2-x+4}{x^2-3 x+2} \\ & =\left(2 x^2+5 x+14\right)+\frac{31 x-24}{(x-2)(x-1)} \quad \text{.....(i)}\end{aligned}$

Let $\frac{31 x-24}{(x-2)(x-1)}=\frac{A}{x-1}+\frac{B}{x-2}$

$\Rightarrow 31-24=-A(x-2)+B(x-1)$ ..... (ii)

Put $x=1$

$\Rightarrow 31-24=-A \Rightarrow A=-7$

Put $x=2$ in Eq. (ii)

$\begin{aligned} & 31 \times 2-24=0+B \Rightarrow B=38 \\ & \therefore \text { From Eq. (i) } f(x)=2 x^2+5 x+14 \\ & \text { and } A+B=-7+38=31 \end{aligned}$

$f^{\prime}(x)=x+\frac{1}{x}$, then integrate it

$f(x)=\int\left(x+\frac{1}{x}\right) d x+c=\frac{x^2}{2}+\log x+c$

$\begin{aligned} & f(x)=\frac{1}{\cos ^2 x \sqrt{1+\tan x}}=\frac{\sec ^2 x}{\sqrt{1+\tan x}} \\ & \Rightarrow \quad \int f(x) d x=\int \frac{\sec ^2 x}{\sqrt{1+\tan x}} d x \\ & \Rightarrow \text { Let } 1+\tan x=u \text {, then } \sec ^2 x d x=d u \\ & \Rightarrow \quad \int f(x) d x=\int \frac{d u}{\sqrt{u}}=2 \sqrt{u}+C \\ & \Rightarrow \quad F(x)=2 \sqrt{1+\tan x}+C \\ & \text { Given, } F(0)=4=2 \sqrt{1}+C \Rightarrow C=2 \\ & \therefore \quad F(x)=2(\sqrt{1+\tan x}+1) \\ & \end{aligned}$

Given,

$\begin{aligned} & \begin{aligned} \int \cos (\log x) d x & =f(x)\{\cos (g(x))+\sin (h(x))\} \\ I & =\int_\limits{\mathrm{II}} \cos (\underset{\mathrm{I}}{\log x) d x} \\ \{ & \text { using by-part method of integration }\} \\ & =\cos (\log x) \cdot x+\int \frac{\sin (\log x)}{x} \cdot x d x \\ & =\cos (\log x) \cdot x+\int \sin (\log x) d x\end{aligned} \\ & =\cos (\log x) \cdot x+\left[\sin (\log x) \cdot x-\int \frac{\cos (\log x)}{x} \cdot x d x\right] \\ & I=x \cdot \cos (\log x)+x \cdot \sin (\log x)-I \\ & 2 I=x(\cos (\log x)+\sin (\log x)) \\ & \Rightarrow I=\frac{x}{2}(\cos (\log x)+\sin (\log x)) \\ & \therefore \quad f(x)=\frac{x}{2} \Rightarrow f^{\prime}(x)=\frac{1}{2}\end{aligned}$

$\begin{aligned} & \quad \int \frac{\sec x}{\sqrt{\sin (2 x+\theta)+\sin \theta}} d x=I \text { (say) } \\ & I=\int \frac{\sec x}{\sqrt{\sin 2 x \cos \theta+\cos 2 x \sin \theta+\sin \theta}} d x \\ & I=\int \frac{\sec x}{\sqrt{\sin 2 x \cos \theta+(\cos 2 x+1) \sin \theta}} d x \\ & =\int \frac{\sec x}{\sqrt{2 \cos x(\sin x \cos \theta+\cos x \sin \theta)}} d x \quad [\because \cos2x+1=2\cos^2x]\\ & =\int \frac{\sec x}{\sqrt{2 \cos ^2 x \cos 2 x \theta(\tan x+\tan \theta)}} d x\end{aligned}$

$\begin{aligned} & =\frac{1}{\sqrt{2 \cos \theta}} \int \frac{\sec ^2 x}{\sqrt{\tan x+\tan \theta}} d x \\ & \text { Let } \tan x+\tan \theta=u \Rightarrow \sec ^2 x d x=d u \\ & =\frac{1}{\sqrt{2 \cos \theta}} \int \frac{d u}{\sqrt{u}}=\frac{1}{\sqrt{2 \cos \theta}} \cdot 2 \cdot \sqrt{u}+c \\ & =\frac{\sqrt{2}}{\sqrt{\cos \theta}} \cdot \sqrt{\tan x+\tan \theta}+c=\sqrt{\frac{2(\tan x+\tan \theta)}{\cos \theta}}+c \\ & =\sqrt{2(\tan x+\tan \theta) \sec \theta}+c\end{aligned}$

$\begin{aligned} & \text { } \frac{3 x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{(x+3)} \\ & \Rightarrow 3 x-2=A(x+1)(x+3)+B(x+3)+C(x+1)^2 \\ & \text { Putting } x=-1 \\ & \qquad-5=2 B \\ & \Rightarrow \quad B=\frac{-5}{2} \\ & \text { Putting } x=-3 \\ & -11=4 C \Rightarrow C=\frac{-11}{4} \end{aligned}$

Putting $x=-3$

$-11=4 C \Rightarrow C=\frac{-11}{4}$

Putting $x=0$

$\begin{aligned} -2 & =3 A+3 B+C \Rightarrow-2=3 A-\frac{15}{2}-\frac{11}{4} \\ \Rightarrow \quad \frac{33}{4} & =3 A \Rightarrow A=\frac{11}{4} \\ \Rightarrow 4 A+2 B & +4 C=11-5-11=-5 \end{aligned}$

$\begin{aligned} \int \frac{\sin \alpha d \alpha}{\sqrt{1+\cos \alpha}} & =\int \frac{\sin \alpha \sqrt{1-\cos \alpha}}{\sqrt{1-\cos ^2 \alpha}} d \alpha \\ & =\int \sqrt{1-\cos \alpha} d \alpha \\ & =\sqrt{2} \int \sin \frac{\alpha}{2} d \alpha \\ & =-2 \sqrt{2} \cos \frac{\alpha}{2}\end{aligned}$

$\int \frac{1+\cos 4 x}{\cot x-\tan x} d x=\int \frac{2 \cos ^2 2 x}{\frac{\cos 2 x}{\cos x \sin x}} d x$

$\begin{aligned} & =\int(2 \cos 2 x \cos x \sin x)=\frac{1}{2} \int 2 \cos 2 x \sin 2 x d x \\ & =\frac{1}{2} \int \sin 4 x d x=\frac{-1}{8} \cos 4 x+C \Rightarrow k=-\frac{1}{8}\end{aligned}$

$\begin{aligned} & \int-\cos x \operatorname{cosec} x \cot x d x \\ & =-\int \cot ^2 x d x \\ & =\int\left(1-\operatorname{cosec}^2 x\right) d x=x+\cot x+C \\ & \therefore \quad f(x)=x \text { and } g(x)=\cot x \\ & f(x) \cdot g(x)=x \cot x \\ & \end{aligned}$

$\begin{aligned} & \int \frac{(2 x+1)^6}{(3 x+2)^8} d x \\ & \int\left(\frac{2 x+1}{3 x+2}\right)^6\left(\frac{1}{3 x+2}\right)^2 d x \\ & \text { Let } \frac{2 x+1}{3 x+2}=t \\ & \Rightarrow \quad \frac{(3 x+2) 2-(2 x+1) 3}{(3 x+2)^2}=\frac{d t}{d x} \Rightarrow\left(\frac{1}{3 x+2}\right)^2 d x=d t \\ & \Rightarrow \quad \int t^6 d t=\frac{t^7}{7}+C \Rightarrow P=\frac{1}{7}, Q=7 \\ & P / Q=1 / 49=1 / 72\end{aligned}$

$\begin{aligned} & \frac{-x^2+6 x+13}{(3 x+5)\left(x^2+4 x+4\right)}=\frac{-x^2+6 x+13}{(3 x+5)(x+2)^2} \\ & \Rightarrow \quad \frac{-x^2+6 x+13}{(3 x+5)(x+2)^2}=\frac{A}{3 x+5}+\frac{B}{x+2}+\frac{C}{(x+2)^2} \\ & \Rightarrow \quad-x^2+6 x+13=A(x+2)^2+B(3 x+5)(x+2)+C(3 x+5) \ldots(\mathrm{i})\end{aligned}$

It is an identity

$\therefore$ True for every value of $x$.

Put $x=-2$ in Eq. (i),

$\begin{aligned} -(-2)^2+6(-2)+13=A \cdot 0+B \cdot 0+C(3(-2)+5) \\ \Rightarrow -4-12+13=-C \Rightarrow C & =3 \end{aligned}$

Put $x=\frac{-5}{3}$ in Eq. (i), we get

$\begin{aligned} -\left(\frac{-5}{3}\right)^2 & +6\left(\frac{-5}{3}\right)+13 \\ =A\left(\frac{-5}{3}+2\right)^2+B \cdot 0+C \cdot 0 \\ \Rightarrow \quad \frac{-25}{9}-10+13 & =A \cdot \frac{1}{9} \\ \Rightarrow \quad \frac{-25}{9}+3 & =\frac{A}{9} \Rightarrow A=2 \end{aligned}$

Put $x=0$ in Eq. (i), we get

$\begin{array}{rlrl} 13 =4 A+10 B+5 C \\ \Rightarrow 13 =8+10 B+15 \\ \Rightarrow 10 B =13-23 \text { or } B=-1 \\ \therefore \frac{-x^2+6 x+13}{(3 x+5)\left(x^2+4 x+4\right)} \\ =\frac{2}{3 x+5}+\frac{-1}{x+2}+\frac{3}{(x+2)^2} \end{array}$

$\begin{aligned} & \text { Let } I=\int \frac{(1+x)+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x \\ & =\int \frac{(1+x)+\sqrt{x} \sqrt{1+x}}{\sqrt{x}+\sqrt{1+x}} d x \\ & =\int \frac{\sqrt{1+x}(\sqrt{1+x}+\sqrt{x})}{(\sqrt{1+x}+\sqrt{x})} d x \\ & =\int \sqrt{1+x} d x\end{aligned}$

Let $1+x=t^2\Rightarrow dx=2tdt$

$\therefore$ $I = 2\int {{t^2}dt = 2{{{t^3}} \over 3} + C = {2 \over 3}{{(1 + x)}^{3/2}} + c} $

$\begin{aligned} & \text { Let } I=\int \frac{(\cos x)}{\mathrm{II}} \frac{\log \cot (x / 2)}{\mathrm{I}} d x \\ & =\log \cot \frac{x}{2} \cdot \int \cos x d x \\ & \quad-\int\left[\frac{d}{d x}\left(\log \cot \frac{x}{2}\right) \int \cos x d x\right] d x\quad [\text {integrating by parts}] \\ & =\log \left(\cot \frac{x}{2}\right) \cdot \sin x-\int \frac{1}{\cot \frac{x}{2}} \\ & \cdot-\operatorname{cosec} \frac{x}{2} \cdot \frac{1}{2} \cdot \sin x d x \\ & =(\sin x) \log \cot \frac{x}{2}+\int 1 d x \\ & =(\sin x) \log \cot \frac{x}{2}+x+c\end{aligned}$

$\begin{aligned} & \text { } I=\int \sqrt{e^{4 x}+e^{2 x} d x}=\int e^x \sqrt{e^{2 x}+1} d x \\ & \text { Put } e^x=t \Rightarrow e^x d x=d t \\ & \therefore \quad I=\int \sqrt{t^2+1} d t \\ & =\frac{1}{2} t \sqrt{t^2+1}+\frac{1}{2.1} \sinh ^{-1}(t)+c \\ & =\frac{1}{2} e^x \sqrt{e^{2 x}+1}+\frac{1}{2} \sinh ^{-1}\left(e^x\right)+c \\ & \end{aligned}$

$\begin{aligned} & \text { } \int \frac{1}{1+\sin x} d x=\tan (f(x))+c \\ & \Rightarrow \int \frac{1-\sin x}{\cos ^2 x} d x=\tan (f(x))+c\end{aligned}$

$\begin{array}{ll}\Rightarrow & \int\left(\sec ^2 x-\sec x \tan x\right) d x=\tan (f(x))+c \\ \Rightarrow & \tan x-\sec x+C=\tan (f(x))+c \\ \Rightarrow & \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=\tan (f(x)) \\ \Rightarrow & -\tan \left(\frac{-\pi}{4}-\frac{x}{2}\right)=\tan (f(x)) \\ \Rightarrow & f(x)=-\frac{\pi}{4}+\frac{x}{2} \Rightarrow f^{\prime}(x)=\frac{1}{2} \\ \text { At } & x=0, f^{\prime}(0)=\frac{1}{2}\end{array}$

$\begin{aligned} & \text { Let } I=\int \frac{e^x(x+3)}{(x+5)^3} d x=\int e^x\left[\frac{1}{(x+5)^2}-\frac{2}{(x+5)^3}\right] d x \\ & \int\left[f(x)+f^{\prime}(x)\right] e^x d x=e^x f(x)+C \\ & \Rightarrow \quad I=\frac{e^x}{(x+5)^2}+C\end{aligned}$

$\begin{aligned} & \text { } \int \frac{(x-1)^2}{\left(x^2+1\right)^2} d x=\int\left(\frac{1}{x^2+1}-\frac{2 x}{\left(x^2+1\right)^2}\right) d x \\ & =\tan ^{-1}(x)-\int \frac{2 x}{\left(x^2+1\right)^2} d x \end{aligned}$

Let $x^2+1=t \Rightarrow 2 x d x=d t$

$\begin{aligned} & \quad=\tan ^{-1}(x)-\int t^{-2} d t=\tan ^{-1}(x)+\frac{1}{t}+k \\ & \Rightarrow \int \frac{(x-1)^2}{(x+1)^2} d x=\tan ^{-1}(x)+\frac{1}{x^2+1}+k \end{aligned}$

Comparing with $\int \frac{(x-1)^2}{\left(x^2+1\right)^2} d x=\tan ^{-1}(x)+g(x)+k$ we have $g(x)=\frac{1}{x^2+1}$

$\begin{aligned} & \text { } \frac{1}{A} \log \left|(\sin x)^{2023}+(\cos x)^{2023}\right|+C \\ & =\int \frac{1-(\cot x)^{2021}}{\tan x+(\cot x)^{2022}} d x=\int \frac{1-\frac{(\cos x)^{2021}}{(\sin x)^{2021}}}{\frac{\sin x}{\cos x}+\frac{(\cos x)^{2022}}{(\sin x)^{2022}}} d x \\ & =\int \frac{(\sin x)^{2021}-(\cos x)^{2021}}{(\sin x)^{2023}+(\cos x)^{2023}} \cdot \cos x \sin x d x\end{aligned}$

$\begin{aligned} & =\int \frac{\cos x(\sin x)^{2022}-\sin x(\cos x)^{2022}}{(\sin x)^{2023}+(\cos x)^{2023}} d x \\ & \because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+C \\ & \Rightarrow \quad \frac{1}{A} \log \left|(\sin x)^{2023}+(\cos x)^{2023}\right|+C \\ & \quad=\frac{1}{2023} \log \left|(\sin x)^{2023}+(\cos x)^{2023}\right|+C \\ & \Rightarrow \quad A=2023\end{aligned}$