Indefinite Integration

$ \begin{aligned} &\text {Given, } \int x^3 \sin 3 x d x\\\\ &\begin{aligned} & \quad=f(x) \cos 3 x+g(x) \sin 3 x+C \\\\ & \text { Let, } I=\int x^3 \sin 3 x d x \\\\ & =x^3 \int \sin 3 x d x-\int\left(3 x^2 x \int \sin 3 x d x\right) d x \\\\ & =x^3\left(\frac{-\cos 3 x}{3}\right)-\int\left(3 x^2 \times\left(\frac{-\cos 3 x}{3}\right)\right) d x \\\\ & =-\frac{x^3}{3} \cos 3 x+\int x^2 \cos 3 x d x \\\\ & =-\frac{x^3}{3} \cos 3 x+\left[x^2\left(\frac{\sin 3 x}{3}\right)-\int\left(2 x \times \frac{\sin 3 x}{3}\right) d x]\right. \\\\ & =-\frac{x^3}{3} \cos 3 x+\frac{x^2}{3} \sin 3 x-\frac{2}{3} \int x \sin 3 x d x \\\\ & =-\frac{x^3}{3} \cos 3 x+\frac{x^2}{3} \sin 3 x-\frac{2}{3} \\\\ & \quad\left[\frac{-x}{3} \cos 3 x+\frac{\sin 3 x}{9}\right]+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\left(-\frac{x^3}{3}+\frac{2 x}{9}\right) \cos 3 x+\left(\frac{x^2}{3}-\frac{2}{27}\right) \sin 3 x+6 \\\\ & \therefore f(x)=-\frac{x^3}{3}+\frac{2 x}{9} \text { and } g(x)=\frac{x^2}{3}-\frac{2}{27} \end{aligned}\\\\ &\text { Now, } 27(f(x)+x g(x))\\\\ &\begin{aligned} & =27\left[\left(-\frac{x^3}{3}+\frac{2 x}{9}\right)+x\left(\frac{x^2}{3}-\frac{2}{27}\right)\right] \\\\ & =27\left[\frac{2 x}{9}-\frac{2}{27} x\right]=6 x-2 x=4 x \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int \frac{d x}{9 \cos ^2 2 x+16 \sin ^2 2 x} \\\\ & =\int \frac{\sec ^2 2 x d x}{9+16 \tan ^2 2 x}=\frac{1}{9} \int \frac{\sec ^2 2 x d x}{1+\left(\frac{4}{3} \tan 2 x\right)^2} \\\\ & =\frac{1}{24} \int \frac{\frac{4}{3} \cdot\left(2 \sec ^2 2 x\right)}{1+\left(\frac{4}{3} \tan 2 x\right)^2} \end{aligned} $

$ \begin{aligned} & =\frac{1}{24} \tan ^{-1}\left(\frac{4}{3} \tan 2 x\right)+C \\\\ & {\left[\because \int \frac{f^{\prime}(x) d x}{1+[f(x)]^2}=\tan ^{-1}(f(x))+C\right]} \end{aligned} $

Let,

$ I=\int \frac{2 \cos 3 x-3 \sin 3 x}{\cos 3 x+2 \sin 3 x} d x $

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

$ \therefore \quad I=\int \frac{2 \cos t-3 \sin t}{\cos t+2 \sin t}\left(\frac{d t}{3}\right) $

Let, $2 \cos t-3 \sin t=A(\cos t+2 \sin t)$

$ +B\left[\frac{d}{d t}(\cos t+2 \sin t)\right] $

$ \begin{aligned} & \Rightarrow 2 \cos t-3 \sin t=A(\cos t+2 \sin t) \\\\ & \quad+B(-\sin t+2 \cos t) \\\\ & \Rightarrow 2=A+2 B \text { and }-3=2 A-B \end{aligned} $

On solving these equations, we get

$ A=-\frac{4}{3} \text { and } B=\frac{7}{5} $

$ \therefore I=\frac{1}{3}\left[\int \frac{-\frac{4}{5}(\cos t+2 \sin t)}{+\frac{7}{5}(-\sin t+2 \cos t)}(\cos t+2 \sin t] d t\right] $

$ \begin{aligned} & =\frac{1}{3}\left[-\frac{4}{5} t+\frac{7}{5} \log |\cos t+2 \sin t|\right]+C \\\\ & =\frac{7}{15} \log |\cos 3 x+2 \sin 3 x|-\frac{4}{5} x+C \\\\ & {[\because t=3 x]} \end{aligned} $

We have,

$ \begin{aligned} & \frac{x^4}{\left(x^2+1\right)(x-2)}=f(x)+\frac{A x+B}{x^2+1}+\frac{C}{x-2} \\\\ & \frac{x^4}{\left(x^2+1\right)(x-2)} \\\\ & \quad f(x)\left[\left(x^2+1\right)(x-2)\right] \\\\ & =\frac{+(A x+B)(x-2)+C\left(x^2+1\right)}{\left(x^2+1\right)(x-2)} \\\\ & \Rightarrow \frac{x^4}{\left(x^2+1\right)(x-2)} \end{aligned} $

$ \begin{aligned} & f(x)\left[x^3-2 x^2+x-2\right] \\\\ = & \frac{+\left[A x^2+(B-2 A) x-2 B\right]+C x^2+C}{\left(x^2+1\right)(x-2)} \end{aligned} $

Here, $f(x)$ will be linear function only as highest power in LHS is 4

So, let $f(x)=p x+q$

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

$\begin{aligned} &(p x+q)\left[x^3-2 x^2+x-2\right] \\\\ &= \frac{+\left[A x^2+(B-2 A) x-2 B+c x^2+C\right]}{\left(x^2+1\right)(x-2)} \\\\ & \Rightarrow \frac{x^4}{\left(x^2+1\right)(x-2)} \\\\ & p x^4+(q-2 p) x^3+(p-2 q+A+C) x^2 \\\\ &= \frac{+(q-2 p+B-2 A) x+(C-2 B-2 q)}{\left(x^2+1\right)(x-2)}\end{aligned}$

On comparing both sides, we get

$ \begin{aligned} & p=1, q-2 p=0 \Rightarrow q=2 \\\\ & p-2 q+A+C=0 \Rightarrow A+C=3 \\\\ & q-2 p+B-2 A=0 \Rightarrow B=2 A \\\\ & C-2 B-2 q=0 \Rightarrow C-2 B=4 \end{aligned} $

From Eqs. (i), (ii) and (iii), we get,

$ A=\frac{-1}{5}, C=\frac{16}{5}, B=\frac{-2}{5}, f(x)=x+2 $

So, $f(14)+2 A-B=14+2+0=16=5 C$

We have,

$ \int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) d x=f(x)+C $

Let

$ \begin{array}{l} I=\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) d x \\\\ =\int\left[\left|\sin \frac{x}{2}-\cos \frac{x}{2}\right|+\left|\sin \frac{x}{2}+\cos \frac{x}{2}\right|\right] d x \\\\ =\int\left[\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)-\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)\right] d x \\\\ \qquad\left[\because \frac{5 \pi}{2} < x < \frac{7 \pi}{2} \Rightarrow \frac{5 \pi}{4} < \frac{x}{2} < \frac{7 \pi}{4}\right] \\\\ =\int\left(\cos \frac{x}{2}-\sin \frac{x}{2}-\sin \frac{x}{2}-\cos \frac{x}{2}\right) d x \\\\ =-2 \int \sin \frac{x}{2} d x=4 \cos \frac{x}{2}+C \end{array} $

So, $ f(x)=4 \cos \frac{x}{2} \Rightarrow f^{\prime}(x)=-2 \sin \frac{x}{2} $

$ f^{\prime}\left(\frac{8 \pi}{3}\right)=-2 \sin \frac{4 \pi}{3}=-2 \times \frac{-\sqrt{3}}{2}=\sqrt{3} $

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

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

Let $3 x+2=t \Rightarrow x d x=\frac{d t}{3}$

$ \begin{aligned} & I=\int \frac{\cos (t-2)}{\cos t} \frac{d t}{3} \\\\ & =\frac{1}{3} \int \frac{\cos 2 \cos t+\sin 2 \sin t}{\cos (t)} d t \\\\ & =\frac{1}{3} \int(\cos 2+\sin 2 \tan t) d t \\\\ & =\frac{1}{3}[(\cos 2) \cdot t+\sin 2 \ln (\sec t)+K] \\\\ & =\frac{1}{3}[\cos 2 \cdot(3 x+2)+\sin 2 \cdot \ln [\sec (3 x+2)]+K \end{aligned} $

$\begin{aligned}= & (\cos 2) x+\frac{2}{3} \cos 2 +\frac{1}{3} \sin 2 \cdot \ln [\sec (3 x+2)]+\frac{K}{3} \end{aligned}$

$=(\cos 2) x+\frac{1}{3} \sin 2 \ln [\sec (3 x+2)]+C$

$\begin{aligned} & \text { Let } I=\int \frac{x^5+x}{x^8+1} d x \\\\ & =\int \frac{x^4\left(x+\frac{1}{x^3}\right)}{x^4\left(x^4+\frac{1}{x^4}\right)} d x=\int \frac{\left(x+\frac{1}{x^3}\right) d x}{\left(x^2\right)^2+\left(\frac{1}{x^2}\right)^2-2+2} \\\\ & =\int \frac{\left(x+\frac{1}{x^3}\right) d x}{\left(x^2-\frac{1}{x^2}\right)^2+(\sqrt{2})^2}\end{aligned}$

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

So, $I=\int \frac{d t / 2}{t^2+(\sqrt{2})^2}=\frac{1}{2} \int \frac{d t}{t^2+\sqrt{2}}$ $=\frac{1}{2}\left[\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)\right]=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left[\frac{x^4-1}{\sqrt{2} x^2}\right]+c$

[put the value of $t=x^2-\frac{1}{x^2}$ ]

To solve the integral $\int \frac{1}{x^{m} \sqrt[m]{x^{m}+1}} \, dx$, we start by making the substitution $u = 1 + \frac{1}{x^m}$. This implies:

$ du = \frac{-m}{x^{m+1}} \, dx $

Rearranging gives:

$ \frac{dx}{x^{m+1}} = -\frac{1}{m} \, du $

Now, rewrite the integral in terms of $u$:

$ \int \frac{1}{x^{m+1} \sqrt[m]{1 + \frac{1}{x^m}}} \, dx = \int -\frac{1}{m} u^{-\frac{1}{m}} \, du $

This integral can be computed by:

$ \int -\frac{1}{m} u^{-\frac{1}{m}} \, du = -\frac{1}{m} \frac{u^{1 - \frac{1}{m}}}{1 - \frac{1}{m}} + C $

Simplifying the integral gives:

$ = -\frac{1}{m-1} u^{1 - \frac{1}{m}} + C $

Substitute back $u = 1 + \frac{1}{x^m}$:

$ = -\frac{1}{m-1} \left(1 + \frac{1}{x^m}\right)^{\frac{m-1}{m}} + C $

Realizing that:

$ \left(1 + \frac{1}{x^m}\right)^{\frac{m-1}{m}} = \left(\frac{x^m + 1}{x^m}\right)^{\frac{m-1}{m}} $

This simplifies to:

$ = -\frac{1}{m-1} \left(\frac{x^m + 1}{x^m}\right)^{\frac{m-1}{m}} + C $

Finally, note that:

$ \left(\frac{x^m + 1}{x^m}\right)^{\frac{m-1}{m}} = \left(\frac{\sqrt[m]{x^m + 1}}{x}\right)^{m-1} $

Thus, the expression becomes:

$ = -\frac{1}{m-1} \left(\frac{\sqrt[m]{x^m + 1}}{x}\right)^{m-1} + C $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \frac{4 x^2+5}{(x-2)^4}=\frac{A}{x-2}+\frac{B}{(x-2)^2} \\ & +\frac{C}{(x-2)^3}+\frac{D}{(x-2)^4} \\ & \Rightarrow 4 x^2+5=A(x-2)^3+B(x-2)^2 \\ & +C(x-2)+D \\ & \Rightarrow 4 x^2+5=A\left(x^3-8-3\left(x^2\right)(2)\right. \\ & +3(x)(4)+B\left(x^2+4-4 x\right) \\ & +C(x-2)+D \\ & \Rightarrow \quad 4 x^2+5=A x^3+x^2(-6 A+B) \\ & +x(12 A-4 B+C) \\ & +(-8 A+4 B-2 C+D) \\ & \Rightarrow A=-6 A+B=4 \\ & \Rightarrow \quad B=4 \\ & 12 A-4 B+C=0 \\ & \Rightarrow \quad 0-16+C=0 \Rightarrow C=16 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { and, }-8 A+4 B-2 C+D=5 \\ & \Rightarrow \quad 0+16-32+D=5 \\ & \Rightarrow \quad D=21 \\ & \therefore \sqrt{\frac{A+B+D}{C}}=\sqrt{\frac{0+4+21}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4} \end{aligned} $

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

Let $x=t^4 \Rightarrow d x=4 t^3 d t$

$ \begin{aligned} I= & \int \frac{t}{\left(t^2+t\right)} \times 4 t^3 d t \\ = & 4 \int \frac{t^4}{t^2+t} d t=4 \int \frac{t^3}{t+1} d t \\ = & 4 \int\left(\left(t^2-t+1\right)-\frac{1}{t+1}\right) d t \\ = & 4\left[\frac{t^3}{3}-\frac{t^2}{2}+t-\log |t+1|\right]+C \\ = & \frac{4}{6}\left[2 \sqrt[4]{x^3}-3 \sqrt[4]{x^2}\right. \\ & \quad+6 \sqrt[4]{x}-6 \log |1+\sqrt[4]{x}|]+C \end{aligned} $

Here, $A=2, B=-3, C=6, D=-6$

$ \begin{aligned} & \frac{2}{3}(A+B+C+D) \\ & =\frac{2}{3}(2-3+6-6)=-\frac{2}{3} \end{aligned} $

Let $I=\int(\log x)^m x^n d x$

Let $\log x=t \Rightarrow \frac{1}{x} d x=d t$ and $x=e^t$

$ I=\int t^m \times\left(e^t\right)^{n+1} d t=\int t^m \times e^{(n+1) t} d t $

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

Let $x=a \sec ^2 \theta$

$ \begin{aligned} & \Rightarrow d x=2 a \sec ^2 \theta \tan \theta d \theta \\ & =\int \sin ^{-1} \sqrt{\frac{a \tan ^2 \theta}{a \sec ^2 \theta}} \times 2 a \sec ^2 \theta \tan \theta d \theta \\ & =\int \theta \times 2 a \sec ^2 \theta \tan \theta d \theta \end{aligned} $

Let, $\tan \theta=t \Rightarrow \sec ^2 \theta d \theta=d t$

$ \begin{aligned} & =2 a \int t \times \tan ^{-1} t d t \\ & =2 a\left[\tan ^{-1} t \times \frac{t^2}{2}-\int \frac{1}{1+t^2} \times \frac{t^2}{2} d t\right] \\ & =2 a\left[\tan ^{-1} t \times \frac{t^2}{2}-\frac{1}{2} \int\left[1-\frac{1}{\left(1+t^2\right)} d t\right]\right. \\ & =2 a\left[\tan ^{-1} t \times \frac{t^2}{2}-\frac{1}{2} t+\frac{1}{2} \tan ^{-1} t\right]+C \\ & =2 a\left[\frac{\tan ^{-1} t}{2}\left[\left(t^2+1\right)\right]-\frac{t}{2}\right]+C \\ & =2 a\left[\frac{\theta}{2}\left(1+\tan ^2 \theta\right)-\frac{\tan \theta}{2}\right]+C \\ & =2 a\left[\sec ^{-1} \sqrt{\frac{x}{a}} \times \frac{x}{a}-\frac{1}{2} \sqrt{\frac{x}{a}-1}\right]+C \\ & =x \cos ^{-1} \sqrt{\frac{a}{x}}-\sqrt{a x-a^2}+C \end{aligned} $

Let $I=\int \frac{\sin x \cos x}{\sqrt{\cos ^4 x-\sin ^4 x}} d x$

Let $\sin ^2 x=t \Rightarrow 2 \sin x \cos x d x=d t$

$ \begin{aligned} & =\frac{1}{2} \int \frac{1}{\sqrt{(1-t)^2-t^2}} d t \\ & =\frac{1}{2} \int \frac{1}{\sqrt{1-2 t}} d t \\ & =\frac{\sqrt{1-2 t}}{-2}+C \\ & =\frac{\sqrt{1-2 \sin ^2 x}}{-2}+C \end{aligned} $

Here, $f(x)=\sqrt{1-2 \sin ^2 x}=\sqrt{\cos 2 x}$

$\therefore$ Domain of $f(x)$ is

$ \left[(4 n-1) \frac{\pi}{4},(4 n+1) \frac{\pi}{4}\right], n=0,1,2, \ldots $

$[\because \cos 2 x$ is positive in (I) and IV quadrant and $\cos (-\theta)=\cos \theta$ ]

Given,

$ \begin{aligned} & \frac{13 x+43}{2 x^2+17 x+30}=\frac{A}{2 x+5}+\frac{B}{x+6} \\ & \Rightarrow \frac{13 x+43}{2 x^2+17 x+30}=\frac{A(x+6)+B(2 x+5)}{(2 x+5)(x+6)} \\ & \Rightarrow 13 x+43=x(A+2 B)+6 A+5 B \end{aligned} $

On compare both sides, we get

$ \begin{array}{lc} & A+2 B=13 \\ \text { and } & 6 A+5 B=43 \\ \Rightarrow \quad & 6 A+12 B=78 \\ & 6 A+5 B=43 \\&- - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \\ \hline & 7 B=35 \\ & B=5 \Rightarrow A=3 \end{array} $

So, $A+B=3+5=8$

$ \begin{aligned} & \text { } \because I=\int e^{4 x^2+8 x-4}(x+1)\cos \left(3 x^2+6 x-4\right) d x \\ & =\int e^{4\left(x^2+2 x\right)-4}(x+1)\left(\cos \left(3\left(x^2+2 x\right)-4\right) d x\right. \\ & \text { Put } x^2+2 x=t \Rightarrow 2(x+1) d x=d t \\ & \therefore \quad I=\frac{1}{2} \int e^{4 t-4} \cos (3 t-4) d t \end{aligned} $

$ \begin{aligned} &\text { We know that, }\\ &\begin{aligned} & \int e^{a x+b} \cos (c x+d) d x \\ & =\frac{e^{a x+b}}{a^2+b^2}(-a \cos (c x+d)+c \sin (c x+d))+c \\ & \therefore I=\frac{1}{2} \frac{e^{4 t-4}}{3^2+(4)^2} [4 \cos (3 t-4)+3 \sin (3 t-4)]+c \\ & =\frac{e^{4 t-4}}{50}[4 \cos (3 t-4)+3 \sin (3 t-4)]+c \\ & =\frac{e^{4 x^2+8 x-4}}{50}\left[4 \cos \left(3 x^2+6 x-4\right)\right. \left.+3 \sin \left(3 x^2+6 x-4\right)\right]+c \end{aligned} \end{aligned} $

$I=\int\left[(\log 2 x)^2+2 \log 2 x\right] d x$

Let $\log 2 x=t \Rightarrow x=e^t / 2$

$ \begin{aligned} & \Rightarrow \frac{1}{2 x} \cdot 2 d x=d t \Rightarrow d x=x d t \Rightarrow d x=\frac{e^t}{2} d t \\ & \therefore \int \frac{1}{2}\left(t^2+2 t\right) e^t d t \\ & \quad=\frac{1}{2}\left[\left(t^2+2 t\right) e^t-\int(2 t+2) e^t d t\right] \\ & \quad=\frac{1}{2}\left[\left(t^2+2 t\right) e^t-\left\{(2 t+2) e^t-\int 2 e^t d t\right\}\right] \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{1}{2}\left[\left(t^2+2 t\right) e^t-(2 t+2) e^t+2 e^t\right] \\ & =\frac{e^t}{2}\left[t^2+2 t-2 t-2+2\right] \\ & =\frac{e^t}{2} t^2+c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned}\\ &\text { On putting value of } t \text { and } e^t / 2 \text { in Eq. (i) }\\ &=x(\log 2 x)^2+c \end{aligned} $

Let

$ I=\int \log \left(6 \sin ^2 x+17 \sin x+12\right) \cos x d x $

Let $\sin x=t$

$ \begin{aligned} & \cos x \cdot d x=d t \\ & =\int \cos x \log \left(6 \sin ^2 x+17 \sin x+12\right) d x \\ & I=\int 1 \cdot \log \left(6 t^2+17 t+12\right) d t \end{aligned} $

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text { II } \,\,\,\,\,\,\,\quad \text { I } $

$ \begin{aligned} & =\log \left(6 t^2+17 t+12\right) \cdot t-\int \frac{12 t+17}{6 t^2+17 t+12} \cdot t d t \\ & =t \log \left(6 t^2+17 t+12\right)-\int \frac{12 t^2+17 t}{6 t^2+17 t+12} d t \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =t \log \left(6 t^2+17 t+12\right) \\ & \quad-\int\left(-\frac{4}{3 t+4}-\frac{3}{2 t+3}+2\right) d t \\ & =t \log \left(6 t^2+17 t+12\right)+\frac{4}{3} \log (3 t+4) +\frac{3}{2} \log (2 t+3)-2 t \end{aligned}\\ &\text { Now, put } t=\sin x\\ &\begin{aligned} & =\sin x \log \left(6 \sin ^2 x+17 \sin x+12\right)+\frac{4}{3} \log \\ & \begin{aligned} (3 \sin x+4) & +\frac{3}{2} \log (2 \sin x+5)-2 \sin x+C \\ \text { Now, } f\left(\frac{\pi}{2}\right) & =1 \cdot \log (6+17+12) \\ & +\frac{4}{3} \log (3+4)+\frac{3}{2} \log (2+3)-2 \\ & =\frac{1}{6}[15 \log 5+14 \log 7-29] \end{aligned} \end{aligned} \end{aligned} $

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

Let $x=\sqrt{2} \tan u$

$ \begin{aligned} d x & =\sqrt{2} \sec ^2 u \cdot d u \\ & =\int \frac{\sqrt{2} \sec ^2 u}{\left(1+2 \tan ^2 u\right) \sqrt{2 \tan ^2 u+2}} d u \\ & =\int \frac{\sqrt{2} \sec ^2 u}{\left(1+2 \tan ^2 u\right) \sqrt{2} \cdot \sec u} d u \\ & =\int \frac{\sec u}{1+2 \tan ^2 u} d u \\ & =\int \frac{\sec u}{1+2 \tan ^2 u} \times \frac{\cos ^2 u}{\cos ^2 u} \cdot d u \\ & =\int \frac{\cos u}{\cos ^2 u+2 \sin ^2 u} d u \end{aligned} $

$ \begin{aligned} &=\int \frac{\cos u}{1+\sin ^2 u} d u\\ &\text { Let } \sin u=t\\ &\begin{aligned} \cos u \cdot d u & =d t=\int \frac{1}{1+t^2} d t \\ & =\tan ^{-1}(t)=\tan ^{-1}(\sin u) \\ & =\tan ^{-1}\left(\sin \left(\tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)\right)\right)+C \\ I & =-\tan ^{-1}\left(\frac{\sqrt{x^2+2}}{|x|}\right)+C \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \because & \int \sin ^4 x \cos ^4 x d x \\ & =\int\left(\sin ^2 x \cos ^2 x\right)^2 d x \\ & =\int\left(\frac{1}{4} \sin ^2 2 x\right)^2 d x=\frac{1}{16} \int \sin ^4 2 x d x \\ & =\frac{1}{16} \int\left(\frac{3}{8}-\frac{\cos 4 x}{2}+\frac{\cos 8 x}{8}\right) d x \\ & =\frac{1}{128} \int(3-4 \cos 4 x+\cos 8 x) d x \\ & =\frac{1}{128}\left(3 x-\sin 4 x+\frac{\sin 8 x}{8}\right)+c \end{aligned} $

From option (b),

$ \begin{aligned} & \frac{1}{256}\left(-\sin ^2 2 x \sin 4 x-\frac{3}{2} \sin 4 x+6 x\right)+c \\ & =\frac{1}{256}\left(\frac{(\cos 4 x-1)}{2} \sin 4 x-\frac{3}{3} \sin 4 x+6 x\right)+c \\ & =\frac{1}{256}\left(\frac{1}{4} \sin 8 x-2 \sin 4 x+6 x\right)+c \\ & =\frac{1}{128}\left(3 x-\sin 4 x+\frac{\sin 8 x}{8}\right)+c \end{aligned} $

Hence, $\int \sin ^4 x \cos ^4 x d x=\frac{1}{256}$

$ \left(-2 \sin ^3 2 x \cos 2 x-3 \sin 2 x \cos 2 x+3\right)+c $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } I=\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x \\ & I=\int \frac{x^2-1}{x^3 \cdot x^2 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} d x \\ & I=\int \frac{\frac{x^2}{x^5}-\frac{1}{x^5}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} d x \\ & \text { Let } 2-\frac{2}{x^2}+\frac{1}{x^4}=t^2 \\ & \text { Then, } \frac{+4}{x^3}-\frac{4}{x^5}=2 t \frac{d t}{d x} \end{aligned} \end{aligned} $

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

$ \begin{aligned} & =\frac{1}{2} \sqrt{\frac{2 x^4-2 x^2+1}{x^4}}+C \\ I & =\frac{\sqrt{2 x^4-2 x^2+1}}{2 x^2}+C \end{aligned} $

$I=\int \frac{x^3 \tan ^{-1} x^4}{1+x^8} d x$

$ \begin{aligned}Let\,\,\,\, x^4 & =t \\ 4 x^3 d x & =d t \\ x^3 d x & =\frac{1}{4} d t \\ I & =\int \frac{\frac{1}{4} \tan ^{-1} t}{1+t^2} d t=\frac{1}{4} \int \frac{\tan ^{-1} t}{1+t^2} d t \end{aligned} $

$ \begin{aligned} & \text { Let } \tan ^{-1} t=z \\ & \frac{1}{1+t^2} d t=d z \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{4} \int z d z \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{4} \cdot \frac{z^2}{2}+C=\frac{z^2}{8}+c \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\left(\tan ^{-1} t\right)^2}{8}+c \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I=\frac{\left(\tan ^{-1}\left(x^4\right)\right)^2}{8}+C \end{aligned} $

$ \begin{aligned} &\text { We have, } \int \frac{2}{x^2+x+1} d x\\ &\begin{aligned} & =2 \int \frac{1}{x^2+x+1} d x \\ & =2 \int \frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} d x \end{aligned}\\ &\text { On substitute } \begin{aligned} u & =\frac{2 x+1}{\sqrt{3}} \rightarrow d u=\frac{2}{\sqrt{3}} d x \\ & =\int \frac{\sqrt{3}}{\left(\frac{3}{4} u^2+\frac{3}{4}\right)} d u \\ & =\frac{4}{\sqrt{3}} \int \frac{1}{u^2+1} d u \\ & =\frac{4}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \end{aligned} \end{aligned} $

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

let $x=\tan \theta$

$ \begin{aligned} d x & =\sec ^2 \theta d \theta \\ I & =\int \frac{\sec ^2 \theta d \theta}{\tan ^2 \theta \sqrt{1+\tan ^2 \theta}} \\ & =\int \frac{\sec ^2 \theta}{\tan ^2 \theta \cdot \sec \theta} d \theta=\int \frac{\frac{1}{\cos \theta}}{\frac{\sin ^2 \theta}{\cos ^2 \theta}} d \theta \\ & =\int \frac{\cos \theta}{\sin \theta} \cdot \frac{1}{\sin \theta} d \theta \end{aligned} $

$ \begin{aligned} & =\int \operatorname{cosec} \theta \cdot \cot \theta d \theta \\ I & =-\operatorname{cosec} \theta+C \\ & =\frac{-\sqrt{1+\tan ^2 \theta}}{\tan \theta}+C \\ I & =-\frac{\sqrt{1+x^2}}{x}+C \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} I & =\int \frac{\sin 7 x}{\sin 2 x \cdot \sin 5 x} d x \\ & =\int\left[\frac{\sin 2 x \cos 5 x}{\sin 2 x \sin 5 x}+\frac{\sin 5 x \cos 2 x}{\sin 2 x \sin 5 x}\right] d x \\ & =\int(\cot 5 x+\cot 2 x) d x \\ & =\frac{1}{5} \log (\sin 5 x)+\frac{1}{2} \log (\sin 2 x)+c \\ & \quad\left[\because \int \cot x d x=\frac{1}{a} \log \sin a x\right] \end{aligned} \end{aligned} $

Given, $\frac{x+2}{\left(x^2+3\right)\left(x^4+x^2\right)\left(x^2+2\right)}$

$ \begin{aligned} &=\frac{A x+B}{x^2+3}+\frac{C x+D}{x^2+2}+\frac{E x^3+F x^2+G x+H}{x^4+x^2} \\ & x+2=\left(x^2+2\right)\left(x^4+x^2\right)(A x+B)+\left(x^2+3\right)\left(x^4+x^2\right)(C x+D)\\ &+\left(E x^3+F x^2+G x+H\right)\left(x^2+2\right)\left(x^2+3\right) \\ &=(A+C+E) x^7+(B+D+F) x^6\\ &+(3 A+4 C+5 E+G) x^5 \\ &+(3 B+4 D+5 F+H) x^4+(2 A \\ &+3 C+6 E+5 G) x^3 \\ &+(2 B+3 D+6 F+5 H) x^2+(6 G) x+6 H \end{aligned} $

On comparing coefficients both sides, we get

$ \begin{array}{r} 6 G=1 \Rightarrow G=\frac{1}{6} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ 6 H=2 \Rightarrow H=\frac{1}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ A+C+E=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ 3 A+4 C+5 E+G=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ 2 A+3 C+6 E+5 G=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ B+D+F=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\\ 3 B+4 D+5 F+H=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vii)\\ 2 B+3 D+6 F+5 H=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(viii) \end{array} $

On putting the value of $G$ in Eq. (iv) and (v) and multiplying 2 by Eq. (iv) and 3 by Eq. (v), we get

On multiply by 3 Eq. (iii) and subtracting from Eq. (iv), we get

$ \text { Now, from Eqs. (ix) and (x), we get } $

Again, from Eqs. (vii) and (viii), we get

$ \begin{aligned} (3 B+4 D+5 F & \left.=-\frac{1}{3}\right) \times 2\left[\because H=\frac{1}{3}\right] \\ (2 B+3 D+6 F & \left.=-\frac{5}{3}\right) \times 3 \\ -D-8 F & =\frac{13}{3} \\ \Rightarrow \quad D+8 F & =-\frac{13}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(xi)\end{aligned} $

On multiply by 3 in Eq. (vi), we get

From Eqs. (xi) and (xii), we get

$ F=-\frac{2}{3} \text { and } D=1 $

On putting the values of $C$ and $E$ in Eq. (iii), we get

$ \begin{aligned} & A+C+E=0 \\ & A+\frac{1}{2}-\frac{1}{3}=0 \Rightarrow A=-\frac{1}{6} \end{aligned} $

On putting the values of $F$ and $D$ in Eq. (vi), we get

$ B=-\frac{1}{3} $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} (E+F) & (C+D)(A) \\ & =\left(-\frac{1}{3}-\frac{2}{3}\right)\left(\frac{1}{2}+1\right)\left(-\frac{1}{6}\right) \\ & =(-1)\left(\frac{3}{2}\right)\left(-\frac{1}{6}\right)=\frac{1}{4} \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} & I=\int \frac{\sin ^6 x}{\cos ^8 x} d x \\ \Rightarrow \quad I & =\int \frac{\sin ^6 x}{\cos ^6 x} \times \frac{1}{\cos ^2 x} d x \\ \Rightarrow \quad I & =\int \tan ^6 x \cdot \sec ^2 x d x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

Let $\tan x=u$

$ \begin{aligned} \Rightarrow \quad \frac{d u}{d x} & =\sec ^2 x \\ d u & =\sec ^2 x d x \end{aligned} $

On putting in Eq. (i), we get

$ \begin{array}{ll} \Rightarrow & I=\int u^6 d u \\ \Rightarrow & I=\frac{u^7}{7}+C \end{array} $

Put the value of $u$

$ I=\frac{\tan ^7 x}{7}+C $

We have,

$ I=\int \frac{x^5}{x^2+1} d x \Rightarrow I=\int \frac{x^4 \cdot x}{x^2+1} $

$ \begin{aligned} & \text { Let } x^2+1=u \Rightarrow x^2=u-1 \\ & \Rightarrow 2 x=\frac{d u}{d x} \Rightarrow \frac{d u}{2}=x d x \text { then, } I \text { become } \\ & \Rightarrow I=\int \frac{(u-1)^2}{u} \cdot \frac{d u}{2} \Rightarrow I=\int \frac{u^2+1-2 u}{u} \frac{du}{2} \\ & \Rightarrow I=\frac{1}{2} \int\left[u+\frac{1}{u}-2\right] d u \end{aligned} $

On integrating, we get

$ I=\frac{1}{2}\left[\frac{u^2}{2}+\log u-2 u\right]+c^{\prime} $

On putting the value of $u$, we get

$ \begin{aligned} & I=\frac{1}{2}\left[\frac{\left(x^2+1\right)^2}{2}+\log \left(x^2+1\right)-2\left(x^2+1\right)\right]+c^{\prime} \\ \Rightarrow \quad & I=\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2} \log \left(x^2+1\right)+c \end{aligned} $

$We\,\,have,\,\,I = \int {\left( {\sum\limits_{r = 0}^\infty {{{{x^r}{3^r}} \over {r!}}} } \right)dx} $

We know that

$ \begin{aligned} & a^m \cdot b^m=(a b)^m \Rightarrow I=\int\left(\sum_{r=0}^{\infty} \frac{(x 3)^r}{r}\right) d x \\ \Rightarrow & I \int\left[\frac{(3 x)^0}{0!}+\frac{(3 x)^1}{1!}+\frac{(3 x)^2}{2!}+\ldots \infty\right] d x \\ \Rightarrow \quad & I=\int\left[1+3 x+(3 x)^2+\ldots \infty\right] d x \end{aligned} $

As, we know that

$ e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\ldots \infty $

Then, $I=\int e^{3 x} d x \Rightarrow I=\frac{e^{3 x}}{3}+C$

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} I & =\int \frac{x^4+1}{x^6+1} d x=\int\left(\frac{1+x^4-x^2}{x^6+1}+\frac{1}{3} \times \frac{3 x^2}{1+x^6}\right) d x \\ & =\int \frac{1}{1+x^2} d x+\frac{1}{3} \int \frac{3 x^2}{1+\left(x^3\right)^2} d x \\ & =\tan ^{-1} x+\frac{1}{3} \tan ^{-1}\left(x^3\right)+C\left[\because 1+x^6=\left(1+x^2\right)\left(1+x^4-x^2\right)\right] \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{array}{ll} \Rightarrow & \quad I=\int e^x(x+1)^2 d x \Rightarrow I=\int e^x\left(x^2+1+2 x\right) d x \\ \Rightarrow & I=\int e^2 x^2 d x+\int e^x d x+2 \int x e^2 d x \\ \Rightarrow & I=x^2 \int e^x d x-\left(\int\left[\frac{d}{d x} x^2\right] \int e^x\right) d x+\int e^x d x +2\left[x \int e^x d x-\int\left(\frac{d}{d x} x \int e^x d x\right) d x\right] \\ \Rightarrow & I=x^2 e^x-2 \int x e^x d x+e^x+2 x e^x-2 \int e^x d x \\ \Rightarrow & I=x^2 e^x-2 x e^x+2 e^x+e^x+2 x e^x-2 e^x+C \\ \Rightarrow \quad & I=e^x\left(x^2+1\right)+C \end{array} \end{aligned} $

$ \text { } \begin{aligned} & \frac{1}{(3 x+1)(x-2)}=\frac{A}{(3 x+1)}+\frac{B}{(x-2)} \\ &=\frac{(x-2) A+B(3 x+1)}{(3 x+1)(x-2)} \\ &=\frac{A x-2 A+3 B x+B}{(3 x+1)(x-2)} \\ & \frac{1}{(3 x+1)(x-2)}=\frac{x(A+3 B)-2 A+B}{(3 x+1)(x-2)} \\ &\,\,\,\,\,\,\, A+3 B=0 \\ &-2 A+B=1 \end{aligned} $

$ \text { } \begin{aligned}\,\,and \frac{(x+1)}{(3 x+1)(x-2)} & =\frac{C}{(3 x+1)}+\frac{D}{(x-2)} =\frac{C(x-2)+D(3 x+1)}{(3 x+1)(x-2)} \\ \frac{(x+1)}{(3 x+1)(x-2)} & =\frac{C x-2 C+3 D x+D}{(3 x+1)(x-2)} \\ \frac{(x+1)}{(3 x+1)(x-2)} & =\frac{x(C+3 D)-2 C+D}{(3 x+1)(x-2} \\ C+3 D & =1 \\ -2 C+D & =1 \end{aligned} $

$ \begin{aligned} &\text { From Eq. (i) } \times 28 \text { and (ii) Eq., we get }\\ &\begin{array}{r} 2 A+6 B=0 \\ -2 A+B=1 \\ \hline 7 B=1 \\ B=\frac{1}{7} \end{array} \end{aligned} $

On putting in Eq. (i), we get

$ A=-3 \times \frac{1}{7}=\frac{-3}{7} $

From Eqs. (iii) and (iv) $\times 3$, we get

$ \begin{gathered} C+3 D=1 \\ \,\,\,\,\,\,-6 C+3 D=3 \\ +\quad-\quad- \\ 7 C=-2 \\ C=-\frac{2}{7} \quad D=\frac{3}{7} \end{gathered} $

$ \begin{aligned} & \qquad A: C=-\frac{3}{7} ;-\frac{2}{7} \Rightarrow 3: 2 \\ & \text { and } B: D=1: 3 \end{aligned} $

Simplify using trigonometric identities.

$ \begin{aligned} & \int \sqrt{1-\sin 2 x d x} \quad[\therefore \sin 2 x=2 \sin x \cos x] \\ & =\int \sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x d x} \\ & \left.\qquad \therefore \therefore 1=\sin ^2 x+\cos ^2 x\right] \\ & =\int \sqrt{(\cos x-\sin x)^2} d x \\ & \int(\cos x-\sin x) d x \\ & =\sin x-(-\cos x)+c=\sin x+\cos x+c \\ & =\cos x+\sin x+c \end{aligned} $

Therefore, the integral of $\sqrt{1-\sin 2 x}$ w.r.t. $x$, if $\sin x+\cos x+c$. where $c$ is the constant of integration.

$ \begin{aligned} & \int e^x \cdot\left(\frac{x+2}{x+4}\right)^2 d x \\ & \quad=\int e^x \frac{x^2+4 x+4}{(x+4)^2} d x \\ & =\int e^x \cdot\left[\frac{x}{x+4}+\frac{4}{(x+4)^2}\right] d x \end{aligned} $

We know that,

$ \begin{aligned} \int e^x(f(x) & \left.+f^{\prime}(x)\right) d x=e^x \cdot f(x)+c \\ & =\int e^x\left[\frac{x}{x+4}+\frac{4}{(x+4)^2}\right] d x \\ & =e^x \cdot \frac{x}{(x+4)}+c \end{aligned} $

Or

(c) Alternate solution

$ \begin{aligned} & \int e^x\left(\frac{x+2}{x+4}\right)^2 d x \\ & \text { Let } u=x+4 \\ & \quad d u=d x ; x=u-4 \\ & \int e^{u-4}\left(\frac{u-4+2}{u}\right)^2 d u \\ & \int e^{u-4}\left(\frac{u-2}{u}\right)^2 d x \Rightarrow \int e^{u-4}\left(1-\frac{2}{u}\right)^2 d u \\ & \int e^{u-4}\left(1+\frac{4}{u^2}-\frac{4}{u}\right) d x \\ & \int\left[e^{u-4}+4 \frac{e^{u-4}}{u^2}-\frac{4 e^{u-4}}{u}\right] d u \\ & e^{-4}\left[e^u-4 \int\left(\frac{e^u}{u}-\frac{e^u}{u^2}\right) d u\right] \end{aligned} $

But Eq. (i)

$ e^{-4}\left[e^u-4 \frac{e^u}{u}\right]+c $

On putting the value of $u=x+4$

$ e^{-4}\left[e^{x+4}-\frac{4 e^{x+4}}{x+4}\right]+c $

$ \begin{aligned} & e^x-\frac{4 e^x}{(x+4)}+c \\ & =e^x\left(\frac{x+4-4}{x+4}\right)+c=\frac{x e^x}{x+4}+c \end{aligned} $

$ \begin{aligned} & \text { If } \int \frac{1}{1-\cos x} d x=\tan \left(\frac{x}{\alpha}+\beta\right)+c \\ & \frac{\pi \alpha}{4}-\beta=\text { ? } \\ & \int \frac{1}{1-\cos x} d x=\tan \left(\frac{x}{\alpha}+\beta\right)+c \end{aligned} $

$ \begin{aligned} & \int \frac{1}{1-1+2 \sin ^2\left(\frac{x}{2}\right)} d x =\tan \left(\frac{x}{\alpha}+\beta\right)+c \\ & \frac{1}{2} \int \operatorname{cosec}^2\left(\frac{x}{2}\right)d x=\tan \left(\frac{x}{\alpha}+\beta\right)+c \left(\because \sin ^2 \frac{x}{2}=\frac{1}{\operatorname{cosec}^2 \frac{x}{2}}\right) \end{aligned} $

$ \begin{aligned} & \frac{1}{2} x-2 \cot \left(\frac{x}{2}\right)=\tan \left(\frac{x}{\alpha}+\beta\right)+c \\ & -\cot \left(\frac{x}{2}\right)=\tan \left(\frac{x}{\alpha}+\beta\right)+c \\ & -\cot \left(\frac{x}{2}\right)=\tan \left(\frac{x}{\alpha}+\beta\right)+c\quad\left(\because \cot y=\frac{1}{\tan y}\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow-\cot \left(\frac{x}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{x}{2}\right) \\ & \Rightarrow \tan \left(\frac{x}{2}-\frac{\pi}{2}\right)+C=\tan \left(\frac{x}{\alpha}+\beta\right)+c \end{aligned} $

On comparison, we get

$ -\frac{\pi}{2}+\frac{x}{2}=\frac{x}{\alpha}+\beta $

On solving these and we get

$ \alpha=2, \beta=-\frac{\pi}{2} $

Now, the value of $\frac{\pi \alpha}{4}-\beta=\pi$

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

$ \begin{aligned} &\text { Let } \quad t=1-\cos ^{(1-n)} \theta\\ &\begin{aligned} & \Rightarrow \quad d t=-(1-n) \cos ^{(-n)} \theta(-\sin \theta) d \theta \\ & \Rightarrow \quad \frac{d t}{1-n}=\cos ^{(-n)} \theta \sin \theta d \theta \end{aligned}\\ &\begin{aligned} \therefore \quad I & =\frac{1}{1-n} \int(t)^{\frac{1}{n}} d t=\frac{1}{1-n} \frac{t^{\frac{1+n}{n}}}{\frac{1+n}{n}}+C \\ & =\frac{n}{1-n^2}\left(1-\cos ^{(1-n)} \theta\right)^{\frac{(n+1)}{n}}+C \end{aligned} \end{aligned} $

Let

$ \begin{aligned} & \frac{x^2+3}{x^4+2 x^2+9}=\frac{x^2+3}{\left(x^2-2 x+3\right)\left(x^2+2 x+3\right)} \\ & =\frac{A x+B}{x^2-2 x+3}+\frac{C x+D}{x^2+2 x+3} \\ & =\frac{\left(x^2+2 x+3\right)(A x+B)+\left(x^2-2 x+3\right)(C x+D)}{\left(x^2-2 x+3\right)\left(x^2+2 x+3\right)} \end{aligned} $

On comparing the numerator, we get

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

Comparing the like terms, we get

$ \begin{aligned} A+C & =0 \\ 2 A+B-2 C+D & =1 \\ 3 A+2 B+3 C-2 D & =0 \\ 3 B+3 D & =3 \end{aligned} $

On solving these equations, we get

$ \begin{aligned} & A=0, B=\frac{1}{2}, C=0 \text { and } D=\frac{1}{2} \\ \therefore \quad & \frac{x^2+3}{\left(x^2-2 x+3\right)\left(x^2+2 x+3\right)} \\ & =\frac{1}{2\left(x^2-2 x+3\right)}+\frac{1}{2\left(x^2+2 x+3\right)} \\ \therefore \quad & a=-2, b=3 \text { and } c=2 \\ & A=0, B=\frac{1}{2}, \\ & C=0, D=\frac{1}{2} \end{aligned} $

Thus, $a A+b B+c C+D$

$ =(-2)(0)+\frac{1}{2}(3)+0+\frac{1}{2}=2 $

$\int \frac{d x}{x\left(x^4+1\right)}$

Let $x^4=t$

then, $4 x^3 d x=d t$

$ \begin{aligned} & \begin{aligned} \int \frac{d x}{x\left(x^4+1\right)} & =\int \frac{d t}{4 x^3 \cdot x(t+1)}=\int \frac{d t}{4 t(t+1)} \\ & =\frac{1}{4} \int\left[\frac{-1}{(t+1)}+\frac{1}{t}\right] d t \\ & =\frac{1}{4}[-\log |t+1|+\log |t|]+C \\ & =\frac{1}{4} \log \left[\frac{t}{t+1}\right]+C \\ & =\frac{1}{4} \log \left|\frac{x^4}{x^4+1}\right|+C, \quad\left(\because x^4=t\right) \end{aligned} \end{aligned} $

$ \begin{aligned} {Let} I & =\int \frac{d x}{\sqrt{\sin ^3 x \cos (x-\alpha)}} \\ & \begin{aligned} \therefore \quad I & =\int \frac{d x}{\sqrt{\sin ^4 x \frac{(\cos x \cos \alpha+\sin x \sin \alpha)}{\sin x}}} \\ & =\int \frac{\operatorname{cosec}^2 x d x}{\sqrt{\cot x \cos ^2+\sin \alpha}} \end{aligned} \end{aligned} $

Let $\cot x=z$

$ \begin{aligned} & \Rightarrow-\operatorname{cosec}^2 x d x=d z \\ &=-\int \frac{d z}{\sqrt{\cos \alpha z+\sin \alpha}} \\ &=-\int \frac{d z}{\sqrt{\cos \alpha(z+\tan \alpha)}} \end{aligned} $

$ =-\frac{1}{\sqrt{\cos \alpha}} \int \frac{d z}{\sqrt{z+\tan \alpha}} $

Let $z+\tan \alpha=t^2$

$ \begin{aligned} \Rightarrow d z & =2 t d t=\frac{-2}{\sqrt{\cos \alpha}} \int \frac{t d t}{\sqrt{t^2}} \\ & =\frac{-2}{\sqrt{\cos \alpha}} \int d t=\frac{-2}{\sqrt{\cos \alpha}} \cdot t+c \\ & =\frac{-2}{\sqrt{\cos \alpha}} \cdot \sqrt{z+\tan \alpha}+c \\ & =\frac{-2 \sqrt{\cot x+\tan \alpha}}{\sqrt{\cos \alpha}}+c \end{aligned} $

Let $I=\int \frac{e^{2 x}}{\sqrt[4]{e^x+1}} d x$

$ \begin{aligned} \text { Let }\left(e^x+1\right)^{\frac{1}{4}} & =t \\ \Rightarrow \quad e^x & =t^4-1 \\ \Rightarrow \quad e^{2 x} & =\left(t^4-1\right)^2 \\ \Rightarrow 2 e^{2 x} \cdot d x & =2\left(t^4-1\right) \cdot 4 t^3 \cdot d t \\ I & =\int \frac{\left(t^4-1\right) \cdot 4 t^3 d t}{t} \\ & =4 \int t^2\left(t^4-1\right) d t \\ & =4\left[\frac{t^7}{7}-\frac{t^3}{3}\right]+C \\ & =4 t^3\left[\frac{t^4}{7}-\frac{1}{3}\right]+C \\ & =\frac{4}{21} t^3\left[3 t^4-1\right]+C \\ & =\frac{4}{21}\left(e^x+1\right)^{\frac{3}{4}}\left(3 e^x-4\right)+C \end{aligned} $

$\int \frac{2-\sin x}{2 \cos x+3} d x$

$ =\int \frac{2}{2 \cos x+3} d x-\int \frac{\sin x}{2 \cos x+3} $

$ =2 \int \frac{\sec ^2\left(\frac{x}{2}\right)}{\tan ^2 \frac{x}{2}+5} d x+\frac{1}{2} \int \frac{1}{u} d u $

$ \left[\begin{array}{l} \text { let } 2 \cos x+3=u \\ \Rightarrow-2 \sin x d x=d \end{array}\right] $

$ =2 \int \frac{2 \sqrt{5}}{5 t^2+5} d t+\frac{1}{2} \log |u| $

$\begin{aligned} & {\left[\text { Let } \frac{1}{\sqrt{5}} \tan \left(\frac{x}{2}\right)=t \Rightarrow \frac{1}{2 \sqrt{5}} \sec ^2 \frac{x}{2} d x=d x\right]} \\ & =\frac{4 \sqrt{5}}{5} \int \frac{d t}{t^2+1}+\frac{1}{2} \log |2 \cos x+3| \\ & =\frac{4}{\sqrt{5}} \tan ^{-1} \frac{t}{1}+\frac{\log |2 \cos x+3|}{2}+C \\ & =\frac{4}{\sqrt{5}} \tan ^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{5}}\right)+\frac{\log (2 \cos x+3)}{2}+C\end{aligned}$

$I=\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$

Let $x=a \tan ^2 t$

$\Rightarrow d x=2 a \tan t \sec ^2 t d t$

$I=\int \sin ^{-1} \sqrt{\frac{a \tan ^2 t}{a+a \tan ^2 t}}\left(2 a \tan t \sec ^2 t\right) d t$

$=2 a \int \sin ^{-1} \sqrt{\frac{a \tan ^2 t}{a \sec ^2 t}} \tan t \sec ^2 t d t$

$=2 a \int t\left(\tan t \sec ^2 t\right) d t$

$=2 a\left[t \int \tan t \sec ^2 t d t-\int \tan t \sec ^2 t d t\right]$

$=2 a\left[\frac{t \tan ^2 t}{2}-\int \frac{\tan ^2 t}{2} d t\right]$

$=a t \tan ^2 t-a \tan t+a t+C$

$=a\left[\frac{x}{a} \tan ^{-1}\left(\sqrt{\frac{x}{a}}\right)-\sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}\right]+C$ $=(a+x) \tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{a x}+C$

We have, $\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)\left(x^2+b^2\right)}$

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

On comparing constants and variables, we get

$ \begin{array}{ll} & A+B=0 \Rightarrow B=-A \\ & C-B a=0 \Rightarrow C-(-A) a=0 \Rightarrow C=-A a \\ & A b^2-C a=1 \\ \Rightarrow & A b^2-(-A a) a=1 \Rightarrow A b^2+A a^2=1 \\ \Rightarrow \quad & A\left(b^2+a^2\right)=1 \Rightarrow A=\frac{1}{a^2+b^2} \end{array} $

Therefore, $C=-A a=\frac{-a}{a^2+b^2}$

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

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

Here, $A=1, B=\frac{1}{4}$ and $C=-\frac{1}{4}$

$ \begin{aligned} & 6 A \rightarrow 7 B-5 C=6+\frac{7}{4}-5\left(-\frac{1}{4}\right) \\ & =6+\frac{7}{4}+\frac{5}{4}=9 \end{aligned} $

$ \begin{aligned} & \text { We have } I=\int \frac{3 x^9+7 x^8}{\left(x^2+2 x+5 x^8\right)^2} d x \\ & =\int \frac{3 x^{-7}+7 x^{-8}}{\left(x^{-6}+2 x^{-7}+5\right)^2} d x\left[\because \text { divide by } x^{16}\right] \end{aligned} $

$ \begin{aligned} &\text { Let } x^{-6}+2 x^{-7}+5=t\\ &\left(-6 x^{-7}-14 x^{-8}\right) d x=d t \end{aligned} $

$ \begin{aligned} \left(3 x^{-7}\right. & \left.+7 x^{-8}\right) d x=-\frac{1}{2} d t \\ I & =-\frac{1}{2} \int \frac{1}{t^2} d t=\frac{1}{2 t}+C \\ & =\frac{1}{2\left(x^{-6}+2 x^{-7}+5\right)}+C \\ & =\frac{x^7}{2\left(x+2+5 x^7\right)}+C \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} & \text { We have } I=\int \frac{\cos x+x \sin x}{x(x+\cos x)} d x \\ & =\int \frac{(x+\cos x)-x(1-\sin x)}{x(x+\cos x)} d x \\ & =\int\left(\frac{1}{x}-\frac{1-\sin x}{x+\cos x}\right) d x \\ & =\log |x|-\log (x+\cos x)+C \\ & =\log \left|\frac{x}{x+\cos x}\right|+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have } I=\int \sqrt{\frac{2}{1+\sin x}} d x\\ &\begin{aligned} & =\int \sqrt{\frac{2}{\frac{1+2 \tan x / 2}{1+\tan ^2 x / 2}} d x} \\ & =\int \sqrt{\frac{2 \sec ^2 x / 2}{1+\tan ^2 x / 2+2 \tan x / 2}} d x \\ & =\sqrt{2} \int \frac{\sec x / 2}{1+\tan x / 2} d x \\ & =\sqrt{2} \int \frac{1}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x \end{aligned} \end{aligned} $

$ \begin{aligned} & =\int \frac{1}{\frac{1}{\sqrt{2}} \cos \frac{x}{2}+\frac{1}{\sqrt{2}} \sin \frac{x}{2}} d x \\ & =\int \frac{1}{\sin \left(\frac{\pi}{4}+\frac{x}{2}\right)} d x=\int \operatorname{cosec}\left(\frac{\pi}{4}+\frac{x}{2}\right) d x \\ & =2 \log \left\lvert\, \operatorname{cosec}\left(\frac{\pi}{4}+\frac{x}{2}\right)-\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)+C\right. \\ & B(x)=\cot \left(\frac{\pi}{4}+\frac{x}{2}\right) \\ & B\left(\frac{\pi}{4}\right)=\cot \left(\frac{\pi}{4}+\frac{\pi}{8}\right)=\cot \frac{3 \pi}{8}=\frac{1}{\sqrt{3+2 \sqrt{2}}} \end{aligned} $

We begin with the integral:

$ I = \int \frac{3}{2 \cos^3 x \sqrt{2 \sin 2x}} \, dx $

Let's simplify this expression:

$ I = \frac{3}{2} \int \frac{\sec^3 x}{\sqrt{4 \sin x \cos x}} \, dx $

This can be further rewritten as:

$ I = \frac{3}{4} \int \frac{\sec^2 x \times \sec^2 x \, dx}{\sqrt{\tan x}} $

Now, make the substitution $\tan x = t^2$. Then, the differential $\sec^2 x \, dx$ becomes:

$ \sec^2 x \, dx = 2t \, dt $

Substituting back, the integral becomes:

$ I = \frac{3}{4} \times 2 \int \frac{1}{t} \times (1 + t^4) t \, dt $

Simplifying further, we have:

$ I = \frac{3}{2} \int (1 + t^4) \, dt $

Evaluating this integral:

$ I = \frac{3}{2} t + \frac{3}{2} \times \frac{t^5}{5} + C $

Substituting back for $ t = \sqrt{\tan x} $:

$ I = \frac{3}{2} (\tan x)^{1/2} + \frac{3}{10} (\tan x)^{5/2} + C $

From this result, we identify that $ B = \frac{1}{2} $ and $ A = \frac{5}{2} $.

To solve for $ B D - A C $ given the equation $\frac{1}{x^4+1}=\frac{A x+B}{x^2+\sqrt{2} x+1}+\frac{C x+D}{x^2-\sqrt{2} x+1}$, we follow these steps:

Starting with:

$ \frac{1}{x^4+1} = \frac{(A x+B)(x^2-\sqrt{2} x+1) + (C x+D)(x^2+\sqrt{2} x+1)}{(x^2+\sqrt{2} x+1)(x^2-\sqrt{2} x+1)} $

Simplify the denominator:

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

Thus, we expand the numerator:

$ Ax^3 - \sqrt{2}Ax^2 + Ax + Bx^2 - \sqrt{2}Bx + B + Cx^3 + \sqrt{2}Cx^2 + Cx + Dx^2 + \sqrt{2}Dx + D $

Collecting like terms:

$ x^3 $ terms: $ A + C = 0 $

$ x^2 $ terms: $ -\sqrt{2}A + B + \sqrt{2}C + D = 0 $

$ x $ terms: $ A - \sqrt{2}B + C + \sqrt{2}D = 0 $

Constant terms: $ B + D = 1 $

From the equations:

$ A + C = 0 $

$-\sqrt{2}A + B + \sqrt{2}C + D = 0 $

$ A - \sqrt{2}B + C + \sqrt{2}D = 0 $

$ B + D = 1 $

From equation 4, if $ B = D = \frac{1}{2} $, substituting into equations, we derive:

From equation 1, $ C = -A $.

Substitute into equation 2:

$ -\sqrt{2}A + B + \sqrt{2}(-A) + D = 0 $

This becomes:

$ -2\sqrt{2}A + \frac{1}{2} + \frac{1}{2} = 0 \implies -2\sqrt{2}A = -1 \implies A = \frac{1}{2\sqrt{2}} $

Then $ C = -\frac{1}{2\sqrt{2}} $.

Now calculate $ B D - A C $:

$ B D - A C = \left(\frac{1}{2} \times \frac{1}{2}\right)-\left(\frac{1}{2\sqrt{2}} \times \left(-\frac{1}{2\sqrt{2}}\right)\right) = \frac{1}{4} + \frac{1}{8} = \frac{3}{8} $

We know,

$ \begin{aligned} & \int\left[x f^{\prime}(x)+f(x)\right] d x=x f(x)+C \\ & \quad \text { We have, } \int \frac{2 x^2 \cos x^2-\sin x^2}{x^2} d x \\ & =\int \frac{x \times \cos x^2 \times 2 x-\sin x^2}{x^2} d x \\ & = \\ & =\int \frac{x\left(\sin x^2\right)^{\prime}-\left(\sin x^2\right)}{x^2} d x \\ & = \\ & =\frac{\sin x^2}{x}+C \end{aligned} $

To evaluate the given integral, we start with:

$ I = \int \frac{\log \left(1+x^4\right)}{x^3} \, dx $

Applying integration by parts, let’s simplify:

$ = \log(1+x^4) \int x^{-3} \, dx - \int \left(\frac{d}{dx}(\log(1+x^4)) \cdot \int x^{-3} \, dx\right) \, dx $

The derivative $\frac{d}{dx} \log(1+x^4) = \frac{4x^3}{1+x^4}$, so the expression becomes:

$ = \frac{\log(1+x^4)}{-2x^2} + \int \frac{2x}{1+x^4} \, dx $

The next part simplifies further:

$ = -\frac{1}{2x^2} \log(1+x^4) + \tan^{-1}(x^2) + C $

Rewriting the logarithm expression:

$ = \frac{1}{2x^2} \log\left(\frac{1}{1+x^4}\right) + \tan^{-1}(x^2) + C $

We identify the parts:

$f(x) = \frac{1}{2x^2}$

$g(x) = 1 + x^4$

$h(x) = x^2$

Now, consider:

$ h(x) \left[f(x) + f\left(\frac{1}{x}\right)\right] = x^2 \left[\frac{1}{2x^2} + \frac{x^2}{2}\right] $

Simplifying this expression:

$ = x^2 \left[\frac{1 + x^4}{2x^2}\right] = \frac{g(x)}{2} $