Indefinite Integration

$ \begin{aligned} &\text {We have, }\\ &\begin{array}{r} \int \frac{\sin x \cdot \sec ^2 x-\tan x \cdot \sin x+\cos x}{(1-\cos 2 x)} d x \\ =\int \frac{\sin x \sec ^2 x-\tan x \cdot \sin x+\cos x}{2 \sin ^2 x} d x \\ =\frac{1}{2}\left[\int \frac{\sin x \sec ^2 x}{\sin ^2 x} d x-\int \frac{\tan x \cdot \sin x}{\sin ^2 x} d x\right. \\ \left.+\int \frac{\cos x}{\sin ^2 x} d x\right] \end{array} \end{aligned} $

$ \begin{array}{r} =\frac{1}{2}\left[\int \frac{1+\tan ^2 x}{\sin x} d x-\int \frac{\tan x}{\sin x} d x\right. \\ \left.+\int \cot x \cdot \operatorname{cosec} x d x\right] \\ =\frac{1}{2}\left[\int \operatorname{cosec} x d x+\int \tan x \cdot \sec x d x\right. \\ \left.-\int \sec x d x+\int \cot x \cdot \operatorname{cosec} x d x\right] \end{array} $

$ \begin{aligned} & =\frac{1}{2}\left[\log \left(\tan \frac{x}{2}\right)+\sec x-\log \right. \\ & \left.\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)-\operatorname{cosec} x\right] \\ & \left.=\frac{1}{2}\left[\sec x-\operatorname{cosec} x-\log \left\lvert\, \frac{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}{\tan \left(\frac{x}{2}\right)}\right.\right]\right]+C \end{aligned} $

We have ,

$ f(x)=\int \frac{16 x^7+5 x^{10}}{\left(x^3+2+3 x^8\right)^2} d x $

and $f(0)=1$

$ \begin{aligned} \Rightarrow \int \frac{16 x^7+5 x^{10}}{x^{16}\left(x^{-5}+2 x^{-8}+3\right)^2} d x & =\int \frac{16 x^{-9}+5 x^{-6}}{\left(x^{-5}+2 x^{-8}+3\right)^2} d x \end{aligned} $

$ \begin{aligned} & \text { Let } x^{-5}+2 x^{-8}+3=d t \\ & \begin{array}{l} \Rightarrow\left(-5 x^{-6}+16 x^{-9}\right) d x=d t \\ \Rightarrow \quad\left(5 x^{-6}+16 x^{-9}\right) d x=-d t \end{array} \\ & \begin{array}{lr} \therefore \quad=\int \frac{-d t}{t^2}=-\int t^{-2} d t=\frac{1}{t}+c \end{array} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad f(x)=\frac{1}{x^{-5}+2 x^{-8}+3}+c \\ & =\frac{x^8}{x^3+2+3 x^8}+c \\ & \text { Since, } f(0)=1 \\ & \Rightarrow \quad c=1 \\ & \therefore \quad f(x)=\frac{x^8}{x^3+2+3 x^8}+1 \\ & \Rightarrow \quad f(-1)=\frac{1}{-1+2+3}+1 \\ & \Rightarrow \quad f(-1)=\frac{1}{4}+1=\frac{5}{4} \end{aligned} $

$ \begin{aligned} & \text { } I=\int \frac{(x+3)}{(x-1)^2(2 x-1)} d x \\ & \text { Let } \frac{x+3}{(2 x-1)(x-1)^2}=\frac{P}{2 x-1}+\frac{Q}{x-1} +\frac{R}{(x-1)^2} \end{aligned} $

$ \begin{aligned} \Rightarrow(x+3)=P(x-1)^2+Q(2 x-1) & (x-1) +R(2 x-1) \end{aligned} $

On putting $x=1, R=4$

On putting $x=1 / 2, P=14$

On putting $x=0, Q=-7$

$ \begin{aligned} & \begin{aligned} & \therefore I=\int\left[\frac{14}{2 x-1}+\frac{-7}{x-1}+\frac{4}{(x-1)^2}\right] d x \\ &=14 \times \frac{1}{2} \log (2 x-1)-7 \log (x-1) +4\left(\frac{-1}{x-1}\right)+k \\ &= \frac{-4}{x-1}+7 \log (2 x-1)-7 \log (x-1)+k \\ & \therefore \quad A=-4, B=7, C=-7 \\ & \Rightarrow \quad A+B+C=-4 \end{aligned} \end{aligned} $

$ \begin{aligned} & \int \frac{1+\cos 8 x}{\tan 2 x-\cot 2 x} d x \\ & =\int \frac{2 \cos ^2 4 x}{\frac{\sin 2 x}{\cos 2 x}-\frac{\cos 2 x}{\sin 2 x}} d x \\ & =\int \frac{2 \cos ^2 4 x}{\frac{\sin ^2 2 x-\cos ^2 2 x}{\cos 2 x \cdot \sin 2 x}} d x \\ & =\int \frac{2 \cos ^2 4 x}{\left(\frac{-\cos 4 x}{\sin 4 x}\right) \times 2} d x \\ & =-\int \cos 4 x \cdot \sin 4 x d x \\ & =\frac{-1}{2} \int \sin 8 x d x=\frac{-1}{2}\left(\frac{-\cos 8 x}{8}\right)+c \\ & =\frac{\cos 8 x}{16}+c \end{aligned} $

$ \begin{aligned} &\text { According to the question, }\\ &\begin{aligned} & f(x)=\frac{1}{16} \text { and } g(x)=8 x \\ \Rightarrow & f\left(\frac{1}{4}\right)=\frac{1}{16} \text { and } g\left(\frac{1}{4}\right)=8 \times \frac{1}{4}=2 \\ \therefore & f\left(\frac{1}{4}\right)+g\left(\frac{1}{4}\right)=\frac{1}{16}+2=\frac{33}{16} \end{aligned} \end{aligned} $

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

On putting $\frac{2 x+1}{5 x+3}=t$

$ \begin{gathered} \Rightarrow 2 x+1=5 t x+3 t \\ \Rightarrow(5 t-2) x=1-3 t \\ x=\frac{1-3 t}{5 t-2} \end{gathered} $

$ \begin{aligned} & \therefore \quad f(t)=\frac{1-3 t}{5 t-2}+2 \\ & =\frac{1-3 t+10 t-4}{5 t-2} \\ & =\frac{7 t-3}{5 t-2} \\ & \begin{array}{l} \therefore \quad f(x)=\frac{7 x-3}{5 x-2} \\ \text { Now, } \int f(x) d x=\int\left(\frac{7 x-3}{5 x-2}\right) d x \\ =7 \int \frac{x}{5 x-2} d x-3 \int \frac{1}{5 x-2} d x \\ =\frac{7}{5} \int \frac{(5 x-2)+2}{(5 x-2)} d x-3 \int \frac{1}{5 x-2} d x \\ =\frac{7}{5} \int\left(1+\frac{2}{5 x-2}\right) d x-3 \int \frac{1}{5 x-2} d x \\ =\frac{7}{5}\left[x+2 \times \frac{1}{5} \log |5 x-2|\right] \\ -3 \times \frac{1}{5} \log |5 x-2|+c \\ =\frac{7}{5} x-\frac{1}{25} \log |5 x-2|+c \end{array} \end{aligned} $

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

On putting $\frac{2 x+3}{3-2 x}=t$

$ \begin{aligned} & \Rightarrow \frac{d x}{(3-2 x)^2}=\frac{d t}{12} \\ & \Rightarrow \quad I=\frac{1}{12} \int \cos (\log t) d t \end{aligned} $

On putting $t=e^k \Rightarrow d t=e^k d k$

$ \begin{aligned} & \therefore I=\frac{1}{12} \int e^k \cdot \cos (k) d k \\ & =\frac{1}{12} \cdot \frac{e^k}{2}[\cos k+\sin k]+c \\ & \Rightarrow I=\frac{\left(\frac{2 x+3}{3-2 x}\right)}{24} \end{aligned} $

$ \begin{aligned} & {\left[\cos \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)+\sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)\right]+c} \\ & \Rightarrow \quad g(x)=\log \left(\frac{2 x+3}{3-2 x}\right) \\ & \quad f(x)=\frac{2 x+3}{3-2 x} \\ & \therefore \quad g(1)=\log (f(1)) \end{aligned} $

$ \begin{aligned} & \text { Given, } \frac{x^2-3 x+2}{(x-4)(x-3)^2} \\ & =\frac{A}{x-4}+\frac{B}{x-3}+\frac{C}{(x-3)^2} \\ & x^2-3 x+2=A(x-3)^2 \\ & +B(x-4)(x-3)+C(x-4) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

This is an identity so, true for every value of $x$.

On putting $x=3$ in Eq. (i), we have

$ C=-2 $

On putting $x=4$ in Eq. (i), we have

$ A=6 $

On putting $x=0$ in Eq. (i) and $A=6$ and $C=-2$, we have $B=-5$

Now, $A+B+C=6-5-2=-1$

Given,

$ \begin{aligned} & \frac{x^2+3}{\left(x^2+1\right)\left(x^2+2\right)}=\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+2} \\ & \Rightarrow x^2+3=\left(x^2+2\right)(A x+B) +\left(x^2+1\right)(C x+D) \\ & \Rightarrow \quad x^2+3=A x^3+B x^2+2 A x +2 B+C x^3+x^2 D+C x+D \\ & \Rightarrow \quad x^2+3=(A+C) x^3+(B+D) x^2 +(2 A+C) x+2 B+D \end{aligned} $

On comparing,

$ \begin{aligned} & A+C=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & B+D=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & 2 A+C=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & 2 B+D=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

From Eqs. (i) and (iii), $A=C=0$

From Eqs. (ii) and (iv), $B=2$ and

$ \begin{aligned} & D=-1 \\ & \therefore \quad A+B+C+D=1 \end{aligned} $

Given,

$ \begin{aligned} & f(x)=\int \frac{2 x^3-3 x^2+4 x-5}{x^2} d x \\ & f(x)=\int 2 x-3+\frac{4}{x}-\frac{5}{x^2} d x \\ & f(x)=2 \cdot \frac{x^2}{2}-3 x+4 \ln x+\frac{5}{x}+c \end{aligned} $

Given, $f(1)=1$

$ \begin{aligned} & f(1)=1-3+4 \ln 1+\frac{5}{1}+c \\ & 1=3+c \Rightarrow c=-2 \\ & f(x)=x^2-3 x+4 \ln x+\frac{5}{x}-2 \\ & f(5)=25-15+4 \\ & \ln 5+\frac{5}{5}-2=9+4 \ln 5 \end{aligned} $

$ \begin{aligned} & \text { If } x>0, x \neq(2 n+1) \frac{\pi}{2} \\ & \int x \sqrt{x}-e^{\log _e(\sec x \tan x)}+\frac{3 x^2-2 x+1}{x^2} d x \\ & =\int\left(x^{\frac{3}{2}}-\sec x \tan x+3-\frac{2}{x}+\frac{1}{x^2}\right) d x \\ & =\frac{x^{5 / 2}}{5 / 2}-\sec x+3 x-2 \ln x-\frac{1}{x}+c \\ & =\frac{2}{5} x^2 \sqrt{x}-\sec x+3 x-2 \ln x-\frac{1}{x}+c \end{aligned} $

Given, $\int(2 x-3) \sqrt{3 x+2} d x$

Put $3 x+2=t \Rightarrow 3 d x=d t$

$ \begin{aligned} & \frac{1}{3} \int\left[2\left(\frac{t-2}{3}\right)-3\right] \sqrt{t} d t=\frac{1}{9} \int(2 t-13) \sqrt{t} d t \\ & \Rightarrow \frac{1}{9} \int 2 t^{3 / 2}-13 \sqrt{t} d t \\ & =\frac{1}{9}\left[2 \cdot \frac{t^{5 / 2}}{5 / 2}-13 \frac{t^{3 / 2}}{3 / 2}\right]+c \\ & =\frac{1}{9}\left[\frac{4}{5} t^2-\frac{26}{3} t\right] \sqrt{t}+c \\ & =\frac{2}{135}\left[6(3 x+2)^2-65(3 x+2)\right] \sqrt{3 x+2}+c \\ & =\frac{2}{135}\left[54 x^2-195 x-106\right] \sqrt{3 x+2}+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 $g(x)=\frac{1}{x^2+1}$

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

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

Given partial fraction of

$ \begin{aligned} & \frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}=\frac{A}{x^2+1}+ \frac{B}{\left(x^2+1\right)^2} +\frac{C}{\left(x^2+1\right)^3} \\ & \therefore x^4+24 x^2+28=A\left(x^2+1\right)^2+B\left(x^2+1\right)+C \end{aligned} $

Now compare the coefficient of $x^4, x^2$ constant term

$ \begin{aligned} & \Rightarrow A=1,24=2 A+B, 28=A+B+C \\ & \Rightarrow B=22, C=28-1-22 \Rightarrow C=5 \end{aligned} $

Now $B-2 A+C=22-2+5=25$

Given that, $\int \frac{x^2}{\left(\sqrt{4-x^2}\right)^3} d x$

Substituting, $x=2 \sin \theta$ and

$ d x=2 \cos \theta d \theta $

Putting in Eq. (i),

$ \begin{aligned} & =\int \frac{(2 \sin \theta)^2}{\left.(\sqrt{4-(2 \sin \theta})^2\right)^3} 2 \cos \theta d \theta \\ & =\int \frac{4 \times 2 \times \sin ^2 \theta \cos \theta d \theta}{\left(\sqrt{4-4 \sin ^2 \theta}\right)^3} \\ & =\int \frac{8 \sin ^2 \theta \cos \theta d \theta}{(2 \cos \theta)^3} \\ & =\int \frac{\sin ^2 \theta \cos \theta}{\cos ^3 \theta} d \theta=\int \tan ^2 \theta d \theta \\ & =\int\left(\sec ^2 \theta-1\right) d \theta=\int \sec ^2 \theta d \theta-\int d \theta \\ & =\tan \theta-\theta+C \end{aligned} $

where, $C$ is integration constant.

$ \because \quad x=2 \sin \theta \text { or } \sin \theta=\frac{x}{2} $

$ \begin{aligned} & \therefore \tan \theta=\frac{x}{\sqrt{4-x^2}} \\ & \text { or } \theta=\tan ^{-1}\left(\frac{x}{\sqrt{4-x^2}}\right) \\ & =\frac{x}{\sqrt{4-x^2}}-\tan ^{-1}\left(\frac{x}{\sqrt{4-x^2}}\right)+C \end{aligned} $

Given that,

$ \begin{equation*} \int \frac{d x}{x \ln (x) \ln ^2(x) \ln ^3(x) \ldots \ln ^m(x)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{equation*} $

Substituting, $\ln x=u$

Differentiating w.r.t ' $x$ ',

$ \frac{1}{x} d x=d u $

Putting in Eq. (i)

$ \int \frac{d u}{u u^2 u^3 \ldots u^m} \text { or } \int \frac{d u}{u^{1+2+3+\ldots+m}} $

$ \begin{aligned} & \text { or } \int \frac{d u}{u^{\frac{m(m+1)}{2}}}\left\{1+2+3++n=\frac{n(n+1)}{2}\right\} \\ & \text { or } \int u^{\frac{-m(m+1)}{2}} d u=\frac{u^{\frac{-m(m+1)}{2}+1}}{\frac{-m(m+1)}{2}+1}+C \\ & {\left[\because \int x^n d x=\frac{x^{n+1}}{n+1}\right]} \\ & =\frac{u^{\frac{(m+2)(1-m)}{2}}}{\frac{(m+2)(1-m)}{2}}+C \end{aligned} $

Now, putting the value of $u$.

$ =\frac{(\ln (x))^{\frac{(m+2)(1-m)}{2}}}{\frac{(m-2)(1-m)}{2}}+C $

Comparing with given form,

$ K=\frac{(m+2)(1-m)}{2} $

and $2 K=(m+2)(1-m)$

Given that,

$ \begin{align*} I_m & =\int x^m \cos n x d x=g(x)-\frac{m(m-1)}{n^2} \\ I_{m-2} & =\int x^{m-2} \cos n x d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

Solving integration by parts

$ =\cos n x \int x^{m-2} d x $

$ \begin{aligned} &\begin{aligned} & \left.-\int\left(\frac{d}{d x} \cos n x\right) \int x^{m-2} d x\right) d x \\ & =\cos n x\left[\frac{x^{m-1}}{m-1}\right]+n \int \sin n x \frac{x^{m-1}}{m-1} d x \\ & I_{m-2}=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \\ & \int x^{m-1} \sin n x d x \end{aligned}\\ &\text { Again integrating by parts method, }\\ &\begin{aligned} & I_{m-2}=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \\ & \left\{\sin n x \int x^{m-1} d x\right. \end{aligned} \end{aligned} $

$ \begin{aligned} & \left.-\int\left(\frac{d}{d x} \sin n x \int x^{m-1} d x\right) d x\right\} \\ = & \frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \\ & \left\{\frac{\sin n x \cdot x^m}{m}-\int n \cdot \cos n x \frac{x^m}{m} d x\right\} \end{aligned} $

$ \begin{aligned} & =\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m-1} \cdot \frac{1}{m} \\ & \left\{x^m \sin n x-n \int x^m \cos n x d x\right\} \\ & =\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)} \end{aligned} $

$ \begin{aligned} & \left(x^m \sin n x-n I_m\right) \\ & I_{m-2}=\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)} x^m \sin n x \end{aligned} $

$ -\frac{n^2}{m(m-1)} I_m $

$ \begin{aligned} &\begin{aligned} & \text { or } \frac{n^2}{m(m-1)} I_m=\frac{x^{m-1} \cos n x}{m-1} \\ & \quad+\frac{n}{m(m-1)} x^m \sin n x-I_{m-2} \\ & \text { or } I_m=\frac{m(m-1)}{n^2} \\ & {\left[\frac{x^{m-1} \cos n x}{m-1}+\frac{n}{m(m-1)} x^m \sin n x-I_{m-2}\right]} \\ & I_m=\frac{m}{n^2} x^{m-1} \cos n x+\frac{x^m \sin n x}{n} \\ & \quad-\frac{m(m-1)}{n^2} I_{m-2} \end{aligned}\\ &\text { On comparing with Eq. (i), we get }\\ &g(x)=\frac{x^m \sin n x}{n}+\frac{m}{n^2} x^{m-1} \cos n x \end{aligned} $

Given that, $I_n=\int \sec ^n x d x$

$ \begin{gathered} f(x)=5 I_6-4 I_4 \\ =5 \int \sec ^6 x d x-4 \int \sec ^4 x d x \\ =\int 5 \sec ^4 x \sec ^2 x d x-\int 4 \sec ^2 x \sec ^2 x d x \\ f(x)=\int\left\{5\left(1+\tan ^2 x\right)^2\right. \\ \left.\quad-4\left(1+\tan ^2 x\right)\right\} \sec ^2 x d x \end{gathered} $

Let $u=\tan x$

Differentiating w.r.t. ' $x$ ', we get

$ \begin{aligned} & \quad d u=\sec ^2 x d x \\ & \therefore f(x)=\int\left\{5\left(1+u^2\right)^2-4\left(1+u^2\right)\right\} d u \\ & =\int\left\{5\left(1+u^4+2 u^2\right)-4\left(1+u^2\right)\right\} d u \\ & =\int\left(5+5 u^4+10 u^2-4-4 u^2\right) d u \\ & =\int\left(5 u^4+6 u^2+1\right) d u=5 \frac{u^5}{5}+6 \frac{u^3}{3}+u \\ & f(x)=u^5+2 u^3+u \\ & f(x)=(\tan x)^5+2(\tan x)^3+\tan x \\ & \text { At } x=\frac{\pi}{4}, f\left(\frac{\pi}{4}\right)=\left(\tan \frac{\pi}{4}\right)^5+2\left(\tan \frac{\pi}{4}\right)^3 +\tan \frac{\pi}{4} \end{aligned} $

$ \begin{aligned} & =(1)^5+2(1)^3+(1) \\ f(\pi / 4) & =4 \end{aligned} $

$ \begin{aligned} &\, \text { We have, } \int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}} \\ \,\,\,\,\,\,& \quad=A \tan ^{-1} \sqrt{f(x)}+C \end{aligned} $

$ \text { } \begin{aligned}Let\,\, I & =\int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}} \\ & =\int \frac{x-1}{x(x+1) \sqrt{x+\frac{1}{x}+1}} d x \end{aligned} $

$ \begin{aligned} & \text { Put } x+\frac{1}{x}+1=t \text { and }\left(1-\frac{1}{x^2}\right) d x=d t \\ & =\int \frac{(x-1)}{x(x+1) \sqrt{t}} \times \frac{x^2}{\left(x^2-1\right)} d t \\ & =\int \frac{x}{(x+1)^2 \sqrt{t}} d t \\ & =\int \frac{d t}{\frac{x^2+2 x+1}{x} \sqrt{t}} \\ & =\int \frac{d t}{\left(1+x+\frac{1}{x}+1\right) \sqrt{t}} \\ & =\int \frac{d t}{(1+t) \sqrt{t}}=2 \tan ^{-1} \sqrt{t}+C \\ & =2 \tan ^{-1} \sqrt{x+\frac{1}{x}+1}+C \\ & \therefore \quad A=2 \Rightarrow f(x)=x+\frac{1}{x}+1 \\ & \,\,\,\,\,\, \quad A=2 \Rightarrow f(-1)=-1-1+1=-1 \\ & \therefore \quad(A, f(-1)=(2,-1) \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } f\left(\frac{2 x+3}{3 x+5}\right)=x+4 \\ & \text { Put, } \frac{2 x+3}{3 x+5}=y \\ & \Rightarrow \quad 2 x+3=3 x y+5 y \\ & \Rightarrow \quad x=\frac{5 y-3}{2-3 y} \\ & \therefore f(y)=\frac{5 y-3}{2-3 y}+4=\frac{5 y-3+8-12 y}{2-3 y} \\ & \Rightarrow f(y)=\frac{7 y-5}{3 x-2} \\ & \therefore f(x)=\frac{7 x-5}{3 x-2} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow f(y)=\frac{7 y-5}{3 x-2} \\ & \therefore f(x)=\frac{7 x-5}{3 x-2} \\ & \Rightarrow \int f(x) d x=\int \frac{7 x-5}{3 x-2} d x \end{aligned}\\ &\text { }\\ &\begin{aligned} Put\,\,3 x-2=t & \Rightarrow x=\frac{t+2}{3} \\ d x & =\frac{d t}{3}=\frac{1}{3} \int \frac{7\left(\frac{t+2}{3}\right)-5}{t} d t \\ & =\frac{1}{9} \int \frac{7 t-1}{t} d t \\ & =\frac{7}{9} \int d t-\frac{1}{3} \int \frac{d t}{t} \\ & =\frac{7}{9} t-\frac{1}{9} \log t+C \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{7}{9}(3 x-2)-\frac{1}{9} \log |3 x-2|+C \\ & =\frac{7}{3} x-\frac{1}{9} \log |3 x-2|+C \\ & \text { Here, } A=\frac{7}{3^{\prime}} B=-\frac{1}{9} \\ & \therefore 3 B-A=3\left(-\frac{1}{9}\right)-\frac{7}{3}=-\frac{8}{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \int e^x\left(\frac{x^2-8 x+19}{(x-1)^5}\right) d x=\frac{e^x(l x+m)}{(x-1)^4}+C \\ & \text { Let } I=\int e^x\left(\frac{x^2-5 x+4-3 x+15}{(x-1)^5}\right) d x \\ & \quad I=\int e^x\left(\frac{(x-4)(x-1)}{(x-1)^5}-3 \frac{(x-5)}{(x-1)^5}\right) d x \\ & \quad I=\int e^x\left(\frac{x-4}{(x-1)^4}-\frac{3(x-5)}{(x-1)^5}\right) d x \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Here, } f(x)=\frac{x-4}{(x-1)^4} \\ & \Rightarrow \quad f(x)=\frac{-3(x-5)}{(x-1)^5} \end{aligned} $

$ \begin{aligned} & \therefore & I & \frac{e^x(x-4)}{(x-1)^4}+C \\ & \therefore & l & =1 m=-4 \\ & \therefore & 4 l+m & =4-4=0 \end{aligned} $