Definite Integration

$ \begin{aligned} I & =\int_1^2\left[x^2\right] d x \\ & =\int_1^{\sqrt{2}} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x+\int_{\sqrt{3}}^2 3 d x \\ & =(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3}) \\ & =\sqrt{2}-1+2 \sqrt{3}-2 \sqrt{2}+6-3 \sqrt{3} \\ & =5-\sqrt{2}-\sqrt{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \lim _{n \rightarrow \infty} & \frac{1}{n^2}\left[e^{1 / n}+2 e^{2 / n}+3 e^{3 / n}+\ldots+2 n e^2\right] \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^2}\left[\sum_{k=1}^{2 n} k e^{k / n}\right] \\ & =\lim _{n \rightarrow \infty} \frac{1}{n}\left[\sum_{k=1}^{2 n}\left(\frac{k}{n}\right) e^{k / n}\right] \\ & =\int_0^2 x e^x d x=\left(x e^x-e^x\right)_0^2 \\ & =\left(2 e^2-e^2\right)-(0-1)=e^2+1 \end{aligned} \end{aligned} $

Given, $m, n, p, q$ be four positive inegers

$ \begin{aligned} & \int_0^{2 \pi} \sin ^m x \cos ^n x d x=4 \int_0^{\pi / 2} \sin ^m x \cos ^n x d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \int_0^{2 \pi} \sin ^p x \cos ^n x d x=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \text { And } \int_0^{2 \pi} \sin ^p x \cos ^q x d x=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \\ & \quad a=m+n+p, b=m+n+q \end{aligned} $

From Eq. (i), we know that

$ \int_0^{2 \pi} f(x) d x=4 \int_0^{\pi / 2} f(x) $

If $f$ is periodic with $\pi$ and symmetric in all four quadrants

$\Rightarrow \quad m$ and $n$ are even

$ \int_0^{2 \pi} \sin ^p x \cos ^n x d x=0 $

$\Rightarrow$ Integrand is odd over symmetric interval $[0,2 \pi]$

Since, $n$ is even so, $p$ is odd also,

$ \begin{aligned} & \int_0^\pi \sin ^p x \cos ^q x d x=0 \\ & \Rightarrow \quad \sin ^p x \cos ^q x \text { is odd about } x=\frac{\pi}{2} \\ & \Rightarrow \quad q \text { is odd } \\ & \begin{aligned} \therefore & a=m+n+p \\ & =\text { even }+ \text { even }+ \text { odd }=\text { odd } \\ \text { And } b & =m+n+p \\ & =\text { even }+ \text { even }+ \text { odd }=\text { odd } \end{aligned} \end{aligned} $

$I=\int_0^2 \sqrt{(x+3)(2-x)} d x$

$ \begin{aligned} & \Rightarrow \quad \int_0^2 \sqrt{-x^2-x+6} d x \\ & \Rightarrow \quad \int_0^2 \sqrt{-\left(x^2+x-6\right)} d x \\ & \Rightarrow \quad \int_0^2 \sqrt{\frac{25}{4}-\left(x+\frac{1}{2}\right)^2} d x \end{aligned} $

Put $u=x+\frac{1}{2} \Rightarrow d u=d x$

When $x=0, u=\frac{1}{2}$ and $x=2$

$ \begin{aligned} & u=2+\frac{1}{2}=\frac{5}{2} \\ & I=\int_{\frac{1}{2}}^{\frac{5}{2}} \sqrt{\left(\frac{5}{2}\right)^2-u^2} d u \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad\left[\frac{u}{2} \sqrt{\frac{25}{4}-u^2}+\frac{\frac{25}{4}}{2} \sin ^{-1}\left(\frac{u}{\frac{5}{2}}\right)\right]_{\frac{1}{2}}^{\frac{5}{2}} \\ & \Rightarrow \quad\left[\frac{u}{2} \sqrt{\frac{25}{4}-u^2}+\frac{25}{8} \sin ^{-1}\left(\frac{2 u}{5}\right)\right]_{\frac{1}{2}}^{\frac{5}{2}} \\ & \Rightarrow \quad\left\{\frac{\frac{5}{2}}{2} \sqrt{\frac{25}{4}-\frac{25}{4}}+\frac{25}{8} \sin ^{-1}(1)\right\} -\left\{\frac{1}{4} \sqrt{\frac{25}{4}-\frac{1}{4}}+\frac{25}{8} \sin ^{-1}\left(\frac{1}{5}\right)\right\} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 0+\frac{25}{8} \cdot \frac{\pi}{2}-\frac{1}{4} \sqrt{6}-\frac{25}{8} \sin ^{-1}\left(\frac{1}{5}\right) \\ & \Rightarrow \quad \frac{25 \pi}{16}-\frac{\sqrt{6}}{4}-\frac{25}{8} \sin ^{-1}\left(\frac{1}{5}\right) \\ & \Rightarrow \quad \frac{25}{8}\left(\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{5}\right)\right)-\frac{\sqrt{6}}{4} \\ & \Rightarrow \quad \frac{25}{8} \cos ^{-1}\left(\frac{1}{5}\right)-\frac{\sqrt{6}}{4} \end{aligned} $

$\int_0^{\frac{\pi}{4}} x^2 \sin 2 x d x, I=\int x^2 \sin 2 x d x$

Put $t=2 x \Rightarrow d t=2 \cdot d x$

So, $I=\int \frac{t^2 \sin t}{8} d t=\frac{1}{8} \int t^2 \sin t d t$

$ \begin{aligned} & =\frac{1}{8}\left[t^2 \cdot(-\cos t)-1 \cdot(-2) \cdot \int t \cos t d t\right] \\ & =\frac{1}{8}\left[-t^2 \cos t+2 \int t \cdot \cos t d t\right] \\ & =\frac{1}{8}\left[-t^2 \cos t+2(t \cdot \sin t-(-\cos t))\right] \\ & =\frac{1}{8}\left[-t^2 \cos t+2 t \sin t+2 \cos t\right] \end{aligned} $

$ \begin{aligned} & =\frac{1}{8}\left[(2 x)^2 \cos (2 x)+2(2 x \cdot \sin (2 x))\right. +2 \cos (2 x)] \\ & =\frac{-x^2 \cos (2 x)+x \sin (2 x)}{2}+\frac{\cos (2 x)}{4} \end{aligned} $

$ \begin{aligned} & \text { So, } \int_0^{\frac{\pi}{4}} x^2 \sin 2 x d x \\ & =\left[\frac{-x^2 \cos (2 x)+x \sin (2 x)}{2}+\frac{\cos (2 x)}{4}\right]_0^{\frac{\pi}{4}} \\ & =\left\{\frac{-\left(\frac{\pi}{4}\right)^2 \cos \left(2 \times \frac{\pi}{4}\right)+\frac{\pi}{4} \sin \left(2 \times \frac{\pi}{4}\right)}{2}\right. \left.+\frac{\cos \left(\frac{2 \pi}{4}\right)}{4}\right\}-\left\{\frac{0}{2}+\frac{\cos (0)}{4}\right\} \\ & \Rightarrow\left(0+\frac{\pi}{8} \cdot 1+\frac{1}{4} \cdot 0\right)-\left(0+\frac{1}{4}\right) \\ & \quad=\frac{\pi}{8}-\frac{1}{4}=\frac{\pi-2}{8} \end{aligned} $

$ \begin{aligned} & \int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x d x \\ & f(x)=\sin ^4 x \cos ^6 x \end{aligned} $

Since, $\sin x$ is an odd function and $\sin ^2 x$ is an even function and $\cos ^6 x$ is an even function.

So, $f(x)$ is an even function.

$ \begin{gathered} \therefore \int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x d x=2 \int_0^{2 \pi} \sin ^4 x \cos ^6 x d x \\ {\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x\right]} \\ =2 \times 2 \int_0^\pi \sin ^4 x \cos ^6 x d x \\ {\left[\because \int_0^{n \pi} f(x) d x=n \int_0^T f(x) d x\right]} \\ =4 \int_0^\pi \sin ^4 x \cos ^6 x d x \\ =4 \times 2 \int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^6 x d x \\ {\left[\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right]} \end{gathered} $

$ \begin{aligned} &=8 \int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^6 x d x\\ &[\because \text { If } m \text { and } n \text { are both even, then }\\ &\int_0^{\frac{\pi}{2}} \sin ^m x \cos ^n x d x= \end{aligned} $

$ \left.\frac{[(m-1)(m-3) \ldots 2 \text { or } 1][(n-1)(n-3) \ldots 2 \text { or } 1]}{[(m+n)(m+n-2) \ldots 2 \text { or } 1]} \times \frac{\pi}{2}\right] $

$ \begin{aligned} & =8 \times \frac{3 \times 1 \times 5 \times 3 \times 1}{10 \times 8 \times 6 \times 4 \times 2} \times \frac{\pi}{2} \\ & =\frac{3 \pi}{64} \end{aligned} $

$ \begin{aligned} & I=\int_0^1 x \sin ^{-1} x \cdot d x \\ & \because \int u v \cdot d x=u \int v d x-\int\left(u^{\prime} \int u d x\right) d x \end{aligned} $

Here, $u=\sin ^{-1} x, v=x$

$ \begin{aligned} &\because I=\left[\sin ^{-1} x \int x \cdot d x-\int\left(\frac{1}{\sqrt{1-x^2}} \int x \cdot d x\right) d x\right]_0^1 \\ & =\left[\sin ^{-1} x \cdot \frac{x^2}{2}-\int \frac{x^2}{2 \sqrt{1-x^2}} d x\right]_0^1 \\ & =\left(\sin ^{-1}(1) \cdot \frac{1}{2}-\sin ^{-1}(0) \cdot 0\right)-\int_0^1 \frac{x^2}{2 \sqrt{1-x^2}} d x \\ & =\frac{\pi}{4}-\frac{1}{2} \int_0^1 \frac{x^2}{\sqrt{1-x^2}} d x \end{aligned} $

Let $x=\sin \theta \Rightarrow d x=\cos \theta \cdot d \theta$ at $x=1 \Rightarrow \theta=\pi / 2$ and $x=0 \Rightarrow \theta=0$

$ \begin{aligned} & =\frac{\pi}{4}-\frac{1}{2} \int_0^{\pi / 2} \frac{\sin ^2 \theta \cdot \cos \theta}{\cos \theta} d \theta \\ & =\frac{\pi}{4}-\frac{1}{2} \int_0^{\pi / 2} \sin ^2 \theta \cdot d \theta \\ & =\frac{\pi}{4}-\frac{1}{2} \int_0^{\pi / 2}\left(\frac{1}{2}-\frac{\cos 2 \theta}{2}\right) d \theta \\ & =\frac{\pi}{4}-\frac{1}{2}\left[\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right]_0^{\pi / 2} \\ & =\frac{\pi}{4}-\frac{1}{2}\left[\frac{\pi}{4}-0-0\right]=\frac{\pi}{4}-\frac{\pi}{8}=\frac{\pi}{8} \end{aligned} $

$\because I=\int_{-\pi / 2}^{\pi / 2} \sin (x-[x]) d x$

for $x \in[-\pi / 2,-1] \Rightarrow[x]=-2$

and $x-[x]=x+2$

for $x \in[-1,0) \Rightarrow[x]=-1$

and $x-[x]=x+1$

for $x \in[0,1] \Rightarrow[x]=0$ and $x-[x]=x$

for $x \in\left[1, \frac{\pi}{2}\right] \Rightarrow x-[x]=x-1$

$ \begin{aligned} \therefore I=\int_{-\pi / 2}^{-1} \sin (x & +2 d x+\int_{-1}^0 \sin (x+1) d x +\int_0^1 \sin x \cdot d x+\int_1^{\pi / 2} \sin (x-1) d x \end{aligned} $

$ \begin{aligned} = & {[-\cos (x+2)]_{-\pi / 2}^1+[-\cos (x+1)]_{-1}^0 } +[-\cos x]_0^1+[-\cos (x-1)]_1^{\pi / 2} \\ = & -\cos (1)+\cos (2-\pi / 2)-\cos (1)+\cos (0) -\cos (1)+\cos (0)-\cos (\pi / 2-1)+\cos (0) \\ = & -3 \cos (1)+3+\cos (-(\pi / 2-2)) -\cos (\pi / 2-1) \\ = & 3-3 \cos (1)+\sin (2)-\sin (1) \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \int_0^2 x^2(2-x)^5 d x \\ & =\int_0^2 x^2\left(32-80 x+80 x^2-40 x^3\right. \left.+10 x^4-x^5\right) d x \\ & =\left[\begin{array}{rl} \frac{32 x^3}{3}-\frac{80 x^4}{4}+\frac{80 x^5}{5}-\frac{40 x^6}{6} \left.+\frac{10 x^7}{7}-\frac{x^8}{8}\right]_0^2 \end{array}\right. \\ & =\left(\frac{32 \times 8}{3}-20 \times 16+16 \times 32-\frac{20}{3} \times 64+\frac{10}{7}\right. \left.\times 128-\frac{256}{8}\right)-0\\ & =\frac{256}{3}+16 \times 12-\frac{1280}{3}+\frac{1280}{7}-32 \\ & =\frac{-1024}{3}+\frac{1280}{7}+160 \\ & =\frac{-7168+3840+3360}{21}=\frac{32}{21} \end{aligned} \end{aligned} $

$\because f(x)=\max \left\{x^3-4, x^4-4\right\}$,

$ \begin{aligned} & g(x)=\min \left\{x^2, x^3\right\} \\ & \text { at } x=-1 \Rightarrow x^2=1, x^3=-1 \\ & \Rightarrow \min \{1,-1\}=-1=(-1)^3 \end{aligned} $

Similarly, at $x=0.5, \Rightarrow x^2=0.25$,

$ \begin{aligned} & x^3=0.125 \\ & \Rightarrow \min \{0.25,0.125\}=0.125=(0.5)^3 \end{aligned} $

So, for $x \in[-1,1] \Rightarrow g(x)=x^3$

Similarly, analysis of $f(x)$

When, $x^3-4=x^4-4$

$ \Rightarrow \quad x^3(x-1)=0 $

So, $x=0,1$

for $x<0 \Rightarrow x^4-4>x^3-4$

for $0x^4-x$

thus, $f(x)=\left\{\begin{array}{lc}x^4-4, & -1 \leq x \leq 0 \\ x^3-4, & 0 < x \leq 1\end{array}\right.$

$ \begin{aligned} & \text { Therefore, } f(x)-g(x) \\ & =\left\{\begin{array}{cc} \left(x^4-4\right)-x^3=x^4-x^3-4, & -1 \leq x \leq 0 \\ \left(x^3-4\right)-x^3=-4, & 0 < x \leq 1 \end{array}\right. \\ & \therefore \int_{-1}^1(f(x)-g(x)) d x=\int_{-1}^0\left(x^4-x^3-4\right) d x +\int_0^1-4 \cdot d x \end{aligned} $

$ \begin{aligned} & =\left[\frac{x^5}{5}-\frac{x^4}{4}-4 x\right]_{-1}^0-4[x]_0^1 \\ & =\left[0-\left(-\frac{1}{5}-\frac{1}{4}+4\right)-4(1)-0\right] \end{aligned} $

$ =-8+\frac{1}{5}+\frac{1}{4}=\frac{-160+5+4}{20}=\frac{-151}{20} $

$ \begin{aligned} & \text { Let } I=\int_0^1 \frac{2 x+5}{x^2+3 x+2} d x \\ & \begin{aligned} \frac{2 x+5}{x^2+3 x+2} & =\frac{A}{(x+1)}+\frac{B}{(x+2)} \\ 2 x+5 & =A(x+2)+B(x+1) \\ 2 & =A+B, 5=2 A+B \end{aligned} \end{aligned} $

Solving we get $A=3, B=-1$

$ \begin{aligned} I & =\int_0^1\left(\frac{3}{x+1}-\frac{1}{x+2}\right) d x \\ & =3[\ln \mid(x+1)]_0^1-[\ln |x+2|]_0^1 \\ & =3 \ln 2-\ln 3+\ln 2 \\ & =4 \ln 2-\ln 3 \\ & =4 \log 2-\log 3=\log \left(\frac{2^4}{3}\right) \\ & =\log \left(\frac{16}{3}\right) \end{aligned} $

Let $I=\int_0^1 x^{\frac{5}{2}}(1-x)^{\frac{3}{2}} d x$

$ \text { } \begin{aligned}Put\,\,\,\, & x=\sin ^2 \theta \\ & d x=2 \sin \theta \cos \theta d \theta \\ \Rightarrow I= & \int_0^{\frac{\pi}{2}} \sin ^5 \theta \cdot\left(\cos ^2 \theta\right)^{\frac{3}{2}} \cdot 2 \sin \theta \cdot \cos \theta d \theta \\ = & 2 \int_0^{\frac{\pi}{2}} \sin ^6 \theta \cdot \cos ^4 \theta d \theta \end{aligned} $

Using Wali's theorem,

$ =2\left(\frac{5 \cdot 3 \cdot 1 \cdot 3 \cdot 1}{10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}\right)=\frac{3 \pi}{256} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} \lim _{n \rightarrow-\infty} \frac{1}{n^2} \cdot \sec ^2 \frac{1}{n^2}+ & \frac{2}{n^2} \sec ^2 \frac{4}{n^2} +\ldots+\frac{1}{n^2} \sec ^2 1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\lim _{n \rightarrow-\infty} \sum_{r=1}^n \frac{r}{n^2} \cdot \sec ^2 \frac{r^2}{n^2} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left(\frac{r}{n}\right) \sec ^2\left(\frac{r}{n}\right)^2 \\ & =\int_0^1 x \cdot \sec ^2 x^2 d x \\ & 1 \leq r \leq n \\ & \frac{1}{n} \leq \frac{r}{n} \leq 1 \\ \because \quad & n \rightarrow \infty \\ & 0 \leq \frac{r}{n}<1 \end{aligned}\\ &\text { Put } \quad x^2=t\\ &\begin{aligned} & \Rightarrow \quad 2 x d x=d t \\ & \int_0^1 \sec ^2 t \frac{d t}{2}=\frac{1}{2}[\tan t]_0^1 \\ & \Rightarrow \quad \frac{\tan 1}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^\pi\left(\sin ^5 x \cos ^3 x+\sin ^4 x\right. \\ & \begin{array}{r} \left.\cos ^4 x+\sin ^3 x \cos ^4 x\right) d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow I=\int_0^\pi\left[\sin ^5(\pi-x) \cos ^3(\pi-x)\right. \\ +\sin ^4(\pi-x) \cos ^4(\pi-x) \\ \left.+\sin ^3(\pi-x) \cdot \cos ^4(\pi-x)\right] d x \\ \Rightarrow I=\int_0^\pi-\sin ^5 x \cos ^3 x+\sin ^4 x \cos ^4 x \\ +\sin ^3 x \cos ^4 x d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} \end{aligned} $

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} & 2 I=2 \int_0^\pi\left(\sin ^4 x \cos ^4 x+\sin ^3 x \cos ^4 x\right) d x \\ & I=\int_0^\pi \frac{\sin ^4 2 x}{16} d x+\int_0^\pi \sin ^3 x \cdot \cos ^4 x d x \end{aligned} $

$ \begin{aligned} & \cos x=t \\ & \Rightarrow-\sin x d x=d t \\ & =\frac{1}{16} \int_0^\pi \frac{(1-\cos 4 x)^2}{4} d x+\int_1^{-1} t^4 \cdot\left(1-t^2\right)(-d t) \\ & =\frac{1}{64} \int_0^\pi\left(1-2 \cos 4 x+\cos ^2 4 x\right) d x \text {. } +\int_1^{-1} t^4\left(t^2-1\right) d t \\ & =\frac{1}{64} \int_0^\pi\left(1-2 \cos 4 x+\frac{1+\cos 8 x}{2}\right) d x +\left(\frac{t^7}{5}-\frac{t^5}{5}\right)_1^{-1} \\ & =\frac{1}{64} \int_0^\pi\left(\frac{3}{2}-2 \cos 4 x+\frac{1}{2} \cos 8 x\right) d x +\left(\frac{2}{5}-\frac{2}{7}\right) \\ & =\frac{3}{128}[x]_0^\pi-\frac{2}{64}\left[\frac{\sin 4 x}{4}\right]_0^\pi+\frac{1}{128}\left[\frac{\sin 8 x}{8}\right]_0^\pi +\frac{4}{35} \\ & \Rightarrow \frac{3 \pi}{128}+\frac{4}{35} \end{aligned} $

$ \begin{aligned} & \text {} \begin{aligned} &I=\int_0^1 \frac{x^4+1}{x^6+1} d x \\ & \text { put } x^6+1=\left(x^2\right)^3+1 \\ &=\left(x^2+1\right)\left(x^4-x^2+1\right) \\ &=\int_0^1 \frac{\left(x^4+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)} d x \\ & \Rightarrow \frac{x^4+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)} \\ &=\frac{A}{x^2+1}+\frac{B x^2+C}{x^4-x^2+1} \\ & x^4+1=A\left(x^4-x^2+1\right) +B x^2\left(x^2+1\right)+C x^2+C \end{aligned} \end{aligned} $

$ \begin{array}{lll} \hline 1=A+B & -A+B+C=0 & A+C=1 \\ \hline B=1-A & A-B-C=0 & C=1-A \\ \hline \end{array} $

$ \begin{aligned} & \therefore \quad A-(1-A)-(1-A)=0 \\ & \Rightarrow \quad A-1+A-1+A=0 \\ & \Rightarrow \quad 3 A=2 \\ & \Rightarrow \quad A=2 / 3 \\ & B=C=1-2 / 3=1 / 3 \\ & I=\frac{2}{3} \int_0^1 \frac{1}{x^2+1} d x-\frac{1}{3} \int \frac{x^2+1}{x^4-x^2+1} d x \\ & =\frac{2}{3}\left[\tan ^{-1} x\right]_0^1-\frac{1}{3} \int_0^1 \frac{1+1 / x^2}{x^2+1 / x^2-1} d x \\ & =\frac{2}{3}\left(\frac{\pi}{4}\right)-\frac{1}{3} \int_0^1 \frac{\left(1+1 / x^2\right)}{\left(x-\frac{1}{x}\right)^2+1} d x \\ & =\frac{\pi}{6}-\frac{1}{3}\left[\tan ^{-1}(x-1 / x)\right]_0^1 \\ & =\frac{\pi}{6}-1 / 3\left(\frac{-\pi}{2}\right) \\ & \Rightarrow \frac{\pi}{6}+\pi / 6=\frac{\pi}{3} \end{aligned} $

$I=\int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x$

Let $u=2 x$, then $d u=2 d x$,

$ \text { so } d x=\frac{1}{2} d u $

When $x=-2 \pi, u=2(-2 \pi)=-4 \pi$

$ \begin{aligned} I & =\int_{-2 \pi}^{2 \pi} \sin ^4(2 x) \cos ^6(2 x) d x \\ & =\frac{1}{2} \int_{-4 \pi}^{4 \pi} \sin ^4(u) \cos ^6(u) d u \end{aligned} $

Since, $\sin (u)$ and $\cos (u)$ are periodic with a period of $2 \pi$,

So, $\sin ^4(u)$ and $\cos ^6(u)$ are also periodic with a period of $2 \pi\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} & =\frac{1}{2} \cdot 2 \int_{-2 \pi}^{2 \pi} \sin ^4(u) \cos ^6(u) d u \\ & =\int_{-2 \pi}^{2 \pi} \sin ^4(u) \cos ^6(u) d u \\ & =\frac{1}{2} \cdot 4 \int_0^{2 \pi} \sin ^4(u) \cos ^6(u) d u \\ & =2 \int_0^{2 \pi} \sin ^4(u) \cos ^6(u) d u \end{aligned} $

Since, the function, $f(u)=\sin ^4(u) \cos ^6(u)$ is an even function.

$ \begin{aligned} & \therefore 2 \int_0^{2 \pi} \sin ^4(u) \cos ^6(u) d u \\ & \quad=2 \cdot 2 \int_0^\pi \sin ^4(u) \cos ^6(u) d u \\ & \quad=4 \int_0^\pi \sin ^4(u) \cos ^6(u) d u \\ & \quad=4 \cdot 2 \int_0^{\pi / 2} \sin ^4(u) \cos ^6(u) d u \\ & \quad=8 \int_0^{\pi / 2} \sin ^4(u) \cos ^6(u) d u \end{aligned} $

We know that, $\int_0^{\pi / 2} \sin ^m(u) \cos ^n(u) d u$

$ =\frac{(m-1)(m-3) \ldots(1 \text { or } 2)(n-1)(n-3)\ldots(1 \text { or } 2)}{(m+n)(m+n-2) \ldots(1 \text { or } 2)} \times \alpha $

where, $\alpha=\frac{\pi}{2}$ if both $m$ and $n$ are even, and $\alpha=1$ othernels.

Here, $m=4, n=6$ both are even.

So, $\int_0^{\pi / 2} \sin ^4(4) \cos ^6(4) d u$

$ \begin{aligned} & =\frac{(4-1)(4-3)(6-1)(6-3)(6-5)}{(4+6)(4+6-2)(4+6-4)} \cdot \frac{\pi}{2} \\ & \quad \begin{array}{c} (4+6-6)(4+6-8) \\ =\frac{3 \cdot 1 \cdot 5 \cdot 3 \cdot 1}{10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}=\frac{45}{3840} \cdot \frac{\pi}{2}=\frac{3}{256} \cdot \frac{\pi}{2} \\ =\frac{3 \pi}{512} \\ \therefore 8 \int_0^{\pi / 2} \sin ^4(u) \cos ^6(u) d u \\ \quad=8 \cdot \frac{3 \pi}{512}=\frac{24 \pi}{512}=\frac{3 \pi}{64} \\ \therefore 8 \int_0^{\frac{\pi}{2}} \sin ^4(u) \cos ^6(u) d u=\frac{3 \pi}{64} \end{array} \end{aligned} $

$ \begin{aligned} & \text { Given, } f(t)=\int_0^t \tan ^{(2 n-1)} x d x, n \in \\ & \text { So, } f(t+\pi)=\int_0^{t+\pi} \tan ^{(2 n-1)} x d x \\ & =\int_0^t \tan ^{(2 n-1)} x d x+\int_t^{t+\pi} \tan ^{(2 n-1)} x d x \\ & \text { Let } u=x-\pi \\ & \Rightarrow d u=d x \end{aligned} $

Limit, when $x=t, u=t-\pi$ and

$ x=t+\pi, u=t $

So, $f(t+\pi)=f(t)+\int_t^t \tan ^{(2 n-1)}(u+\pi) d u$

$ \begin{aligned} & \text { Since, } \tan (u+\pi)=\tan u \\ & =f(t)+\int_{t-\pi}^t \tan ^{(2 n-1}(u) d u \\ & =f(t)+\int_{t-\pi}^0 \tan ^{(2 n-1)}(u) d u +\int_0^t \tan ^{(2 n-1)}(u) d u \end{aligned} $

Let $v=-u$, the $d v=-d u$

When $u=t-\pi, v=\pi-t$, when $u=0$, $v=0$, so we get

$ \begin{aligned} =f(t)-\int_{\pi-t}^0 \tan ^{(2 n-1)}(-v) d v & +\int_0^t \tan ^{(2 n-1)}(u) d u \\ =f(t)+\int_{\pi-t}^0 \tan ^{(2 n-1)}(v) d v & +\int_0^t \tan ^{(2 n-1)}(u) d u \end{aligned} $

$ \begin{aligned} & =f(t)+\int_{\pi-t}^0 \tan ^{(2 n-1)}(v) d v+f(t) \\ & =2 f(t)-\int_0^{\pi-t} \tan ^{(2 n-1)}(v) d v \\ & =2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v-\int_{\pi-t}^\pi \tan ^{(2 n-1)}(v) d v \\ & \text { Let } w=v-\pi, \text { then } d w=d v \\ & \text { When } v=\pi-t, w=-t, \text { when } v=\pi \\ & w=0 \\ & =2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v -\int_{-t}^0 \tan ^{(2 n-1)}(w+\pi) d w \\ & 2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v -\int_{-t}^0 \tan ^{(2 n-1)}(w) d w \end{aligned} $

Let $z=-w$, then $d z=-d w$ when $w=-t$, $z=t$ and $w=0, z=0$

$ \begin{aligned} = & 2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v +\int_t^0 \tan ^{(2 n-1)}(-z) d z \\ = & 2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v -\int_0^t \tan ^{(2 n-1)}(z) d z \\ = & 2 f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v-f(t) \\ = & f(t)-\int_0^\pi \tan ^{(2 n-1)}(v) d v \\ = & f(t)+0\left[\because \int_0^\pi \tan ^{(2 n-1)}(v) d v=0\right] \\ & =f(t)+f(\pi) \quad[\because f(\pi)=0] \\ \therefore \quad & f(t+\pi)=f(t)+f(\pi) \end{aligned} $

$ \begin{aligned} I & =\int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x \\ & =\int_0^2 x^8\left(\frac{4-x^2}{x^2}\right)^{5 / 2} d x \\ & =\int_0^2 x^8 \cdot \frac{\left(4-x^2\right)^{5 / 2}}{x^5} d x \\ & =\int_0^2 x^3\left(4-x^2\right)^{5 / 2} d x \end{aligned} $

Let $u=4-x^2$

Then, $d u=-2 x d x$, so $x d x=\frac{-1}{2} d u x^2=4-u$ and $x^3=x \cdot x^2=x(4-u)$ when $x=0, u=4$ and when $x=2, u=0$

$ \begin{aligned} & \therefore \int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x \\ & \quad=\int_0^2 x^3\left(4-x^2\right)^{5 / 2} d x \\ & \quad=\int_4^0(4-u) \cdot u^{5 / 2} \cdot\left(\frac{-1}{2}\right) d u \\ & \quad=\frac{1}{2}\left[4 \cdot \frac{u^{7 / 2}}{7 / 2}-\frac{u^{9 / 2}}{9 / 2}\right]_0^4 \\ & \quad=\frac{1}{2}\left[\frac{8}{7} u^{7 / 2}-\frac{2}{9} u^{9 / 2}\right]_0^4 \\ & \quad=\frac{1}{2}\left[\frac{8}{7}(u)^{7 / 2}-\frac{2}{9}(4)^{9 / 2}-0\right] \\ & \quad=\frac{1}{2}\left[\frac{8}{7}(128)-\frac{2}{9}(512)\right] \\ & \quad=\frac{1}{2}\left[\frac{1024}{7}-\frac{1024}{9}\right] \\ & \quad=\frac{1024}{2} \cdot \frac{2}{63}=\frac{1024}{63} \\ & \therefore \int_0^2 x^8\left(\frac{4}{x^2}-1\right)^{5 / 2} d x=\frac{1024}{63}=\frac{2^{10}}{63} \end{aligned} $

$ \begin{aligned} & \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cdot \cos ^2 x(\sin x+\cos x) d x \\ & =\int_{-\pi 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x \\ & \quad+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x \cos ^3 x d x \\ & =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin x \cdot\left(1-\cos ^2 x\right) \cos ^2 x d x \\ & \quad+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x\left(1-\sin ^2 x\right) \cdot \sin ^2 x d x \end{aligned} $

$ \begin{gathered} =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin x \cdot \cos ^2 x \cdot d x \\ -\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin x \cdot \cos ^4 x d x \\ +\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x \cdot \sin ^2 x \cdot d x-\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos x \cdot \sin ^4 x d x \\ =\left[\frac{\cos ^3 x}{3}+\frac{\cos ^5 x}{5}+\frac{\sin ^3 x}{3}-\frac{\sin ^5 x}{5}\right]_{-\frac{\pi}{2}}^\pi \end{gathered} $

$ \begin{aligned} & =\left(0+0+\frac{1}{3}-\frac{1}{5}\right)-\left(0+0-\frac{1}{3}+\frac{1}{5}\right) \\ & =\frac{2}{3}-\frac{2}{5}=\frac{10-6}{15}=\frac{4}{15} \end{aligned} $

$ \int_{\frac{1}{5}}^{\frac{1}{2}} \frac{\sqrt{x-x^2}}{x^3} d x $

$ =\int_{\frac{1}{5}}^{\frac{1}{2}} \frac{x \sqrt{\frac{1}{x}-1}}{x^3} d x \int {_{{1 \over 5}}^{{1 \over 2}}} {1 \over {{x^2}}}\sqrt {{1 \over x} - 1.dx}$

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

$ \begin{aligned} & =-\int_4^1 t^{\frac{1}{2}} \cdot d t=\int_1^4 \sqrt{t} \cdot d t \\ & =\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_1^4=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_1^4 \\ & =\frac{2}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right]=\frac{2}{3}(8-1) \\ & =\frac{14}{3} \end{aligned} $

$ \begin{aligned} I & =\int_0^{400 \pi} \sqrt{1-\cos 2 x} d x \\ & =\int_0^{400 \pi} \sqrt{2 \sin ^2 x} d x \\ & =\sqrt{2} \int_0^{400 \pi} \sin x d x \\ & =400 \sqrt{2} \int_0^\pi \sin x d x \\ & =2 \times 400 \sqrt{2} \int_0^{\pi / 2} \sin x d x \\ & =800 \sqrt{2}[-\cos x]_0^{\frac{\pi}{2}} \\ & =800 \sqrt{2}\left[-\cos \frac{\pi}{2}+\cos 0\right] \\ & =800 \sqrt{2} \end{aligned} $

Let $I=\int_0^x \frac{t^2}{\sqrt{a^2+t^2}} d t$

Put $t=a \sin h \theta \Rightarrow d t=a \cos h \theta d \theta$

Now, $\sqrt{a^2+t^2}=\sqrt{a^2+a^2 \sin h^2 \theta}$

$ =a \sqrt{1+\sinh ^2 \theta}=a \cos h \theta $

and   $ t^2=a^2 \sin h^2 \theta $

$ \begin{aligned} &\begin{array}{lll} \hline t & 0 & x \\ \hline \theta & 0 & \sin h^{-1}\left(\frac{x}{4}\right) \\ \hline \end{array}\\ &\begin{aligned} & \therefore I=\int_0^{\sin h^{-1}\left(\frac{x}{a}\right)} \frac{a^2 \sin h^2 \theta}{a \cos h \theta} \cdot a \cos h \theta d \theta \\ & \quad=\int_0^{\sin h^{-1}\left(\frac{x}{a}\right)} a^2 \sin h^2 \theta d \theta \\ & \because \sin h^2 \theta=\frac{\cos h(2 \theta)-1}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { So, } I=a^2 \int_0^{\sin h^{-1}\left(\frac{x}{a}\right)} \frac{\cos h(2 \theta)-1}{2} d \theta=\frac{a^2}{2} \\ & {\left[\begin{array}{c} \sin h^{-1}\left(\frac{x}{a}\right) \\ \int_0^{\cos h(2 \theta) d \theta-\int_0^{\sin h^{-1}\left(\frac{x}{a}\right)}} \int_1 1 \cdot d \theta \end{array}\right]} \\ & =\frac{a^2}{2}\left[\frac{1}{2} \sin h(2 \theta)-\theta\right]=\frac{a^2}{2} \\ & =\frac{a^2}{2}\left[\frac{x}{a} \times \frac{\sqrt{a^2+x^2}}{a}-\sin h^{-1}\left(\frac{x}{a}\right)\right] \\ & =\frac{x}{2} \sqrt{a^2+x^2}-\frac{a^2}{2} \sin h^{-1}\left(\frac{x}{a}\right) \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos ^4 x d x \\ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\cos ^4 x} d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^4 x d x \\ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^2 x\left(1+\tan ^2 x\right) d x=\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\left(1+u^2\right) d u \end{aligned} $

$ \begin{array}{lll} \hline \boldsymbol{x} & \frac{\pi}{6} & \frac{\pi}{3} \\ \hline u & \frac{1}{\sqrt{3}} & \sqrt{3} \\ \hline \end{array} $

$ \begin{aligned} &\text { where } u=\tan x\\ &\begin{aligned} & =\left(u+\frac{u^3}{3}\right)_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \\ & =\left(\sqrt{3}+\frac{(\sqrt{3})^3}{3}\right)-\left(\frac{1}{\sqrt{3}}+\frac{\left(\frac{1}{\sqrt{3}}\right)^3}{3}\right) \\ & =\frac{3 \sqrt{3}+3 \sqrt{3}}{3}-\left(\frac{1}{\sqrt{3}}+\frac{1}{9 \sqrt{3}}\right) \\ & =\frac{6 \sqrt{3}}{3}-\left(\frac{9+1}{9 \sqrt{3}}\right)=2 \sqrt{3}-\frac{10 \sqrt{3}}{27} \\ & =\frac{54 \sqrt{3}-10 \sqrt{3}}{27}=\frac{44 \sqrt{3}}{27}=\frac{44}{9 \sqrt{3}} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} Let\,\,& I=\int_0^{\frac{3 \pi}{2}} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \quad \begin{aligned} I= & \int_0^{\frac{3 \pi}{2}} \frac{\left(\cos \left(\frac{3 \pi}{2}-x\right)\right)^3}{\left(\cos \left(\frac{3 \pi}{2}-x\right)\right)^3+\left(\sin \left(\frac{3 \pi}{2}-x\right)\right)^3} \end{aligned} d x \\ &=\int_0^{\frac{3 \pi}{2}} \frac{-\sin ^3 x}{-\sin ^3 x-\cos ^3 x} d x \\ &=\int_0^{\frac{3 \pi}{2}} \frac{\sin ^3 x}{\sin ^3 x+\cos ^3 x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \begin{aligned} &\text { On adding in Eqs. (i) and (ii), we get }\\ &\begin{array}{ll} \Rightarrow & 2 I=\int_0^{\frac{3 \pi}{2}} d x=\frac{3 \pi}{2} \\ \therefore & I=\frac{3 \pi}{4} \end{array} \end{aligned} $

$ \begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\ldots+\frac{1}{k n}\right] \\ & \sum_{i=1}^{(k-1) n} \frac{1}{n+i} \\ & \Rightarrow \frac{1}{n} \sum_{i=1}^{(k-1) n} \frac{1}{1+\frac{i}{n}}=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{(k-1) n} \frac{1}{1+\frac{i}{n}} \\ & =\int_0^{k-1} \frac{1}{1+x} d x=[\ln (1+x)]_0^{k-1} \\ & =\ln k \end{aligned} $

$ \begin{aligned} & \text { } \int_{-1}^4 \sqrt{\frac{4-x}{x+1}} d x \\ & \text { Let } x=\frac{5}{2} \sin \theta+\frac{3}{2} \\ & \Rightarrow d x=\frac{5}{2} \cos \theta d \theta \\ & \Rightarrow 4-x=4-\left(\frac{5}{2} \sin \theta+\frac{3}{2}\right) \\ & =\frac{5}{2}(1-\sin \theta) \\ & \Rightarrow x+1=\frac{5}{2} \sin \theta+\frac{3}{2}+1=\frac{5}{2}(1+\sin \theta) \\ & \Rightarrow \quad \frac{4-x}{x+1}=\frac{1-\sin \theta}{1+\sin \theta} \\ & \Rightarrow \quad \sqrt{\frac{4-x}{x+1}}=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{1-\sin \theta}{\cos \theta} \end{aligned} $

$ \begin{aligned} & \int \sqrt{\frac{4-x}{x+1}} d x=\int \frac{1-\sin \theta}{\cos \theta} \frac{5}{2} \cos \theta d \theta \\ & \quad=\frac{5}{2} \int(1-\sin \theta) d \theta \\ & \quad=\frac{5}{2}(\theta+\cos \theta)+c \\ & x=\frac{5}{2} \sin \theta+\frac{3}{2} \Rightarrow \sin \theta=\frac{2 x-3}{5} \end{aligned} $

$ \begin{gathered} \Rightarrow \cos \theta=\sqrt{1-\sin ^2 \theta}=\frac{2}{5} \sqrt{4+3 x-x^2} \\ \begin{aligned} \int_{-1}^4 \sqrt{\frac{4-x}{x+1}} d x= & {\left[\frac{5}{2} \sin ^{-1}\left(\frac{2 x-3}{5}\right)\right.}+ \left.\sqrt{4+3 x-x^2}\right]_{-1}^4 \end{aligned} \\ =\frac{5}{2} \cdot \frac{\pi}{2}-\frac{5}{2}\left(\frac{-\pi}{2}\right) \\ =\frac{5 \pi}{4}+\frac{5 \pi}{4}=\frac{5 \pi}{2} \end{gathered} $

$ \text { } \begin{aligned} & \int_0^{\frac{\pi}{4}} \frac{\cos ^2 x d x}{\cos ^2 x+4 \sin ^2 x} \\ & \quad=\int_0^{\frac{\pi}{4}} \frac{1}{1+4 \tan ^2 x} d x \end{aligned} $

Put $u=2 \tan x$

$ \begin{aligned} d u & =2 \sec ^2 x d x \\ d x & =\frac{d u}{2\left(1+\tan ^2 x\right)}=\frac{2 d u}{4+u^2} \\ x & =0, u=0 \\ x & =\frac{\pi}{4}, u=2 \\ & =\int_0^2 \frac{1}{1+u^2} \cdot \frac{2 d u}{4+u^2} \\ & =\int_0^2\left(\frac{2}{3} \frac{1}{1+u^2}-\frac{2}{3} \frac{1}{4+u^2}\right) d u \\ & =\frac{2}{3}\left[\tan ^{-1} u\right]_0^2-\frac{2}{3} \cdot \frac{1}{2}\left[\tan ^{-1} \frac{u}{2}\right]_0^2 \\ & =\frac{2}{3} \tan ^{-1}(2)-\frac{\pi}{12} \end{aligned} $

$ \begin{aligned} & \text { } \int_{5 \pi}^{25 \pi} \sin 2 x+\cos 2 x \mid d x \\ & \Rightarrow|\sin 2 x+\cos 2 x|=\sqrt{2}\left|\sin \left(2 x+\frac{\pi}{4}\right)\right| \\ & \Rightarrow \int_{5 \pi}^{25 \pi} \sin 2 x+\cos 2 x \mid d x \\ & =\int_{5 \pi}^{25 \pi} \sqrt{2}\left|\sin \left(2 x+\frac{\pi}{4}\right)\right| d x \end{aligned} $

Let $u=2 x+\frac{\pi}{4}$

$ \begin{aligned} & d u=2 d x \\ & d x=\frac{1}{2} d u \\ & x=5 \pi, u=10 \pi+\frac{\pi}{4} \\ & x=25 \pi, u=50 \pi+\frac{\pi}{4} \\ & =\sqrt{2} \int_{10 \pi+\frac{\pi}{4}}^{50 \pi+\frac{\pi}{4}}|\sin u| \frac{1}{2} d u \\ & =\frac{\sqrt{2}}{2} \int_{10 \pi+\frac{\pi}{4}}^{50 \pi+\frac{\pi}{4}}|\sin u| d u \end{aligned} $

The period of $|\sin u|$ is $\pi$.

The length of the interval of integration is

$ \begin{aligned} & =\left(50 \pi+\frac{\pi}{4}\right)-\left(10 \pi+\frac{\pi}{4}\right)=40 \pi \\ & =\frac{\sqrt{2}}{2} \cdot 40 \int_0^\pi|\sin u| d u \\ & =\frac{\sqrt{2}}{2} \times 40 \times 2 \\ & =40 \sqrt{2} \end{aligned} $

$I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{3}}\left|\tan \left(x-\frac{\pi}{6}\right)\right| d x$

Put $\tan \left(x-\frac{\pi}{6}\right)=0 \Rightarrow x-\frac{\pi}{6}=0$

$ \begin{aligned} & I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{6}}-\tan \left(x-\frac{\pi}{6}\right) d x+\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \tan \left(x-\frac{\pi}{6}\right) d x \\ & \begin{array}{r} =-\left[\log \left(\sec \left(x-\frac{\pi}{6}\right)\right]_{-\pi}^{\frac{\pi}{6}}\right. +\left[\log \sec \left(x-\frac{\pi}{6}\right)\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \end{array} \\ & \Rightarrow \log \left(\sec \left(\frac{5 \pi}{12}\right)\right)+\log \left(\sec \left(\frac{\pi}{6}\right)\right) \\ & \quad=\log \left(\sec \left(\frac{5 \pi}{12}\right) \sec \left(\frac{\pi}{6}\right)\right) \\ & =\log \left(2 \sqrt{2} \frac{(\sqrt{3}+1)}{\sqrt{3}}\right) \end{aligned} $

$ \begin{aligned} \text { } I= & \int_0^\pi \frac{x \sin x}{\sin ^2 x+2 \cos ^2 x} d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow I= & \int_0^\pi \frac{(\pi-x) \sin x}{\sin ^2 x+2 \cos ^2 x} d x \\ & \quad\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right] \end{aligned} $

$ \begin{aligned} & \Rightarrow 2 I=\int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x \\ & \text { Put } \cos x=t \Rightarrow-\sin x d x=d t \\ & \Rightarrow 2 I=-\pi\left[\tan ^{-1} t\right]_1^{-1} \\ & \Rightarrow \quad 2 I=-\pi\left[\frac{-\pi}{4}-\frac{\pi}{4}\right] \Rightarrow I=\frac{\pi^2}{4} \end{aligned} $

$ \text { } \mathop {\lim }\limits_{n \to \infty }\left[\frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\frac{n}{n^2+n^2}\right] $

$ \begin{aligned} & \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r}{r^2+n^2}=\lim _{n \rightarrow \infty} \sum \frac{\left(\frac{r}{n}\right)}{1+\left(\frac{r}{n}\right)^2} \cdot \frac{1}{n} \\ & \Rightarrow \int_0^1 \frac{x}{1+x^2} d x \\ & \text { put } 1+x^2=t \Rightarrow 2 x d x=d t \\ & \Rightarrow \frac{1}{2} \int_1^2 \frac{1}{t} d t \Rightarrow \frac{1}{2}[\log t]_1^2=\frac{1}{2} \log 2 \end{aligned} $

$ \begin{aligned} & \int_0^{\frac{\pi}{2}} \ln (\tan x+\cot x) d x=\int_0^{\frac{\pi}{2}} \ln \left(\frac{1}{\sin x \cos x}\right) d x \\ & =\int_0^{\frac{\pi}{2}}(\ln 2-\ln \sin 2 x) d x \\ & =\ln 2 \int_0^{\frac{\pi}{2}} d x-\int_0^{\frac{\pi}{2}} \ln \sin 2 x d x \\ & \text { put } 2 x=t, d x=\frac{d t}{2} \\ & =\frac{\pi}{2} \ln 2-\frac{1}{2} \int_0^\pi \ln \sin t d t \\ & =\frac{\pi}{2} \ln 2-\int_0^{\pi / 2} \ln \sin t d t \\ & =\frac{\pi}{2} \ln 2-\left(-\frac{\pi}{2} \ln 2\right) \end{aligned} $

$ \left[\because \int_0^{\pi / 2} \ln \sin x=-\frac{\pi}{2} \ln 2\right] $

$ =\pi \ln 2 $

Let $I=\int_0^\pi x \sin ^5 x \cos ^6 x d x$

$ \begin{aligned} &\begin{aligned} \Rightarrow \quad I & =\frac{\pi}{2} \int_0^\pi \sin ^5 x \cos ^6 x d x \\ & =\frac{\pi}{2} 2 \int_0^{\pi / 2} \sin ^5 x \cos ^6 x d x \end{aligned}\\ &\text { By Walli's Formula, }\\ &=\pi\left[\frac{(4 \cdot 2)(5 \cdot 3 \cdot 1)}{11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1}\right]=\frac{8 \pi}{693} \end{aligned} $

$ \begin{aligned} & \text { } I=\int_{1 / 2}^{1 / \sqrt{2}} \frac{1}{\left(x+\sqrt{1-x^2}\right)\left(1-x^2\right)} d x \\ & I=\int_{\pi / 6}^{\pi / 4} \frac{\cos \theta d \theta}{(\sin \theta+\cos \theta) \cos ^2 \theta} \\ & \text { Put } x=\sin \theta \Rightarrow d x=\cos \theta d \theta \\ & \quad=\int_{\pi / 6}^{\pi / 4} \frac{d \theta}{\sin \theta \cos \theta+\cos ^2 \theta} \end{aligned} $

Dividing $N^T$ and $D^r$ by $\cos ^2 \theta$, we get

$ =\int_{\pi / 6}^{\pi / 4} \frac{\sec ^2 \theta}{\tan \theta+1} d \theta $

Put $\tan \theta+1=t, \sec ^2 \theta d \theta=d t$

$ \begin{aligned} & =\int_{\frac{1}{\sqrt{3}}+1}^2 \frac{d t}{t}=[\ln t]_{1+\frac{1}{\sqrt{3}}}^2 \\ & =\ln 2-\ln \left(1+\frac{1}{\sqrt{3}}\right) \\ & =\ln 2-\ln \left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)=\ln \left(\frac{2 \sqrt{3}}{\sqrt{3}+1}\right) \\ & =\ln \left(\frac{2 \sqrt{3}(\sqrt{3}-1)}{3-1}\right)=\ln (3-\sqrt{3}) \end{aligned} $

$ \begin{aligned} &\text { Given, } H(x)=3 x^4+6 x^3-2 x^2+1\\ &\begin{aligned} & \text { Given, } \frac{H(x)}{(x-1)(x-2)(x+1)} \\ & \begin{aligned} &=f(x)+\frac{g(x)}{(x-1)(x+1)(x-2} \text { taking LCM } \\ & H(x)=f(x)(x-1)(x+1)(x-2)+g(x) \\ & H(-1)=g(-1) \\ & H(2)=g(2) \\ & H(1)=g(1) \\ & \text { Now, } H(-1)+2 H(2)-3 H(1) \\ &=g(-1)+2 g(2-3 g(1) \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \begin{aligned} I & =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\cos x-\sin x}{\sin 2 x} d x \\ I & =\int_{\sqrt{2}}^{\frac{\sqrt{3}}{2}+\frac{1}{2}} \frac{d t}{t^2-1} \end{aligned}\\ &\text { Put } \sin x+\cos x=t\\ &\begin{aligned} & (\cos x-\sin x) d x=d t \\ & 1+\sin 2 x=t^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & =\left[\frac{1}{2} \ln \left(\frac{t-1}{t+1}\right)\right]_{\sqrt{2}}^{\frac{\sqrt{3}+1}{2}} \\ & =\frac{1}{2}\left[\ln \left(\frac{\frac{\sqrt{3}+1}{2}-1}{\frac{\sqrt{3}+1}{2}+1}\right)-\ln \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\right] \\ & =\frac{1}{2}\left[\ln \frac{\sqrt{3}-1}{\sqrt{3}+3}-\ln \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\right] \\ & =\frac{1}{2}\left[\ln \frac{2(2 \sqrt{3}-3)}{6}-\ln (3-2 \sqrt{2})\right] \\ & =\frac{1}{2}\left[\ln \frac{2 \sqrt{3}-3}{3}+\ln (3+2 \sqrt{2})\right] \\ & =\frac{1}{2}\left[\ln \left(\frac{2-\sqrt{3}}{\sqrt{3}}\right)+\ln (3+2 \sqrt{2})\right] \\ & =\frac{1}{2} \ln \left[\frac{(3+2 \sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right] \end{aligned} $

$ \text { } \begin{aligned} I & =\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos x+\sin x} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x \end{aligned} $

$ \begin{aligned} & =\frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\sin \frac{x}{2}+\cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} d x \\ & =\frac{1}{2} \int_0^{\frac{\pi}{2}}\left(1+\frac{\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) d x \\ & =\frac{1}{2}\left[x+\left\{-2 \ln \left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\right\}\right]_0^{\frac{\pi}{2}} \\ & =\frac{1}{2}\left[x-2 \ln \left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\right]_0^{\frac{\pi}{2}} \\ & =\frac{1}{2}\left[\left(\frac{\pi}{2}-2 \ln \sqrt{2}\right)-0\right] \\ & =\frac{\pi}{4}-\ln \sqrt{2}=\frac{\pi}{4}-\frac{1}{2} \ln 2 \end{aligned} $

$I=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Apply property

$ \begin{aligned} & \int_a^b x f(x) d x=\frac{a+b}{2} \int f(x) d x \\ & \text { If } f(a+b-x)=f(x) \\ & \qquad I=\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x \end{aligned} $

$ \begin{aligned} & \text { Put } \cos x=t \\ & \sin x d x=-d t \\ & \qquad \begin{aligned} I & =-\frac{\pi}{2} \int_{+1}^{-1} \frac{d t}{1+t^2} \\ & =\frac{\pi}{2} \int_{-1}^1 \frac{d t}{1+t^2}=\frac{\pi}{2} \cdot 2 \int_0^1 \frac{d t}{1+t^2} \\ & =\pi\left(\tan ^{-1} t\right)_0^1 \\ & =\pi \tan ^{-1} 1=\pi \times \frac{\pi}{4}=\frac{\pi^2}{4} \end{aligned} \end{aligned} $

$\begin{aligned} & \int_\limits{\frac{1}{4}}^{\frac{3}{4}} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x \\ & x=\cos 2 \theta \\ & \Rightarrow d x=(-2 \sin 2 \theta \mathrm{d} \theta) \end{aligned}$

Take limit as $\alpha$ and $\beta$

$\begin{aligned} & -2 \int_\limits\alpha^\beta \cos 2 \theta \cdot \sin 2 \theta d \theta \\ & =\int_\limits\alpha^\beta \sin 4 \theta d \theta \\ & =\left.\frac{-\cos 4 \theta}{4}\right|_\alpha ^\beta \\ & =-\left.\frac{1}{4}\left(2 \cdot\left(x^2\right)-1\right)\right|_{1 / 4} ^{3 / 4} \\ & =-\left.\frac{1}{4}\left(2 x^2-1\right)\right|_{1 / 4} ^{3 / 4} \\ & =-\frac{1}{4}\left(\frac{18}{16}-1-\frac{2}{16}+1\right) \\ & =-\frac{1}{4} \end{aligned}$

$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$

Using Newton Leibniz theorem

$\begin{aligned} & =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{\sin \left(2 \times \frac{\pi}{2}\right) \cdot 0-\sin (2 x) \cdot 3 x^2+\left(\cos \frac{\pi}{2}\right) \cdot 0-\cos x \cdot 3 x^2}{2\left(x-\frac{\pi}{2}\right)}\right] \\ & =\lim _\limits w{x \rightarrow \frac{\pi}{2}} \frac{-3 x^2 \sin 2 x-3 x^2 \cos x}{2\left(x-\frac{\pi}{2}\right)}\left(\frac{0}{0}\right) \text { form } \\ & =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{-6 x \sin 2 x-6 x^2 \cos 2 x-6 x \cos x+3 x^2 \sin x}{2}\right] \\ & =\frac{6 \times \frac{\pi^2}{4}+3 \times \frac{\pi^2}{4}}{2} \\ & =\frac{9 \pi^2}{8} \end{aligned}$

$\begin{aligned} & \int_\limits{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x \\ & =\left[x \log _e\left(x+\sqrt{x^2+1}\right)\right]_{-1}^2-\int_\limits{-1}^2 \frac{x}{\left(x+\sqrt{x^2+1}\right)}\left(1+\frac{x}{\sqrt{x^2+1}}\right) d x \\ & =2 \log _2(2+\sqrt{5})+\log _e(\sqrt{2}-1)-\int_\limits{-1}^2 \frac{x}{\sqrt{x^2+1}} d x \\ & =\log _e\left[(2+\sqrt{5})^2(\sqrt{2}-1)\right]-\left[\sqrt{x^2+1}\right]_{-1}^2 \\ & =\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right]-\sqrt{5}+\sqrt{2} \\ & =\sqrt{2}-\sqrt{5}+\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right] \end{aligned}$

$\begin{aligned} & \text { Let } \sqrt{e^x-1}=t \\ & e^x-1=t^2 \\ & e^x=1+t^2 \\ & e^x=0+2 t-\frac{d t}{d x} \\ & \frac{d t}{d x}=\frac{e^x}{2 t}=\frac{t^2+1}{2 t} \\ & I=\int \frac{2 t}{t\left(1+t^2\right)} d t=2 \tan ^{-1} t \\ & \Rightarrow \quad I=\int_\limits\alpha^{\log ^4} \frac{d x}{\sqrt{e^x-1}} \\ & I=\left.2 \tan ^{-1} \sqrt{e^x-1}\right|_\alpha ^{\log 4} \\ & =2\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{e^\alpha-1}\right)=\frac{\pi}{6} \\ & \Rightarrow \frac{\pi}{3}-\tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{12} \\ & \tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{4} \\ & \Rightarrow e^\alpha-1=1 \\ & e^\alpha=2 \Rightarrow e^{-\alpha}=\frac{1}{2} \end{aligned}$

$\therefore \quad$ Quadratic equation whose roots are $e^a$ & $e^{-\alpha}$ is

$\begin{aligned} & x^2-\left(e^\alpha+e^{-\alpha}\right) x+e^\alpha \times e^{-\alpha}=0 \\ & x^2-\left(2+\frac{1}{2}\right) x+1=0 \\ & 2 x^2-5 x+2=0 \end{aligned}$

$\begin{aligned} & I(21)=\int_\limits0^1\left(1-x^k\right)^{21} d x \\ & =\int_\limits0^1\left(1-x^k\right)\left(1-x^k\right)^{20} d x \\ & =\int_\limits0^1\left(1-x^k\right)^{20} d x-\int_0 x^k\left(1-x^k\right)^{20} d x \\ & I(21)=I(20)-\int_\limits0^1 x^k\left(1-x^k\right)^{20} d x \end{aligned}$

$I(21)=I(20)-\left\lfloor\frac{\left(1-x^k\right)^{21}}{-21 k} x-\int_\limits0^1 \frac{(1-x^k)^{21}}{-21 k} d x\right\rfloor$

$\begin{aligned} & I(21)=I(20)-\frac{1}{21 k} I(20) \\ & \Rightarrow[I(21)](21 k+1)=21 K I(20) \\ & \Rightarrow 21 K=147 \Rightarrow K=7 \end{aligned}$

$\begin{aligned} & \int_\limits0^{\pi / 4} \frac{\cos ^2 x \cdot \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \\ & =\int_\limits0^{\pi / 4} \frac{\tan ^2 x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x \end{aligned}$

Let $\tan x=t$

$\int_\limits0^1 \frac{t^2 d t}{\left(1+t^3\right)^2}$

Let $1+t^3=\mathrm{z}$

$\begin{gathered} 3 t^2 d t=d z \\ \frac{1}{3} \int_\limits1^2 \frac{d z}{z^2}=\left.\frac{1}{3}\left(-\frac{1}{z}\right)\right|_1 ^2 \\ =-\frac{1}{3}\left(\frac{1}{2}-1\right)=\frac{1}{6}\end{gathered}$

First, let's rewrite the given integral using the given form of the Beta function. The given integral is:

$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x$

To use the Beta function, let us make a substitution. Let $ x^{10} = t $. Then, $ dx = \frac{1}{10}t^{-\frac{9}{10}} dt $ or $ dx = \frac{1}{10} t^{-\frac{9}{10}} dt $. The limits of integration change as follows: when $ x = 0 $, $ t = 0 $, and when $ x = 1 $, $ t = 1 $.

Substituting these into the integral, we have:

$\int_\limits0^1 (1 - t)^{20} \cdot \frac{1}{10} t^{-\frac{9}{10}} dt$

which simplifies to:

$\frac{1}{10} \int_\limits0^1 (1 - t)^{20} t^{-\frac{9}{10}} dt$

We recognize this integral as a Beta function $ \beta(m, n) $ where $ m = 1 - \frac{9}{10} = \frac{1}{10} $ and $ n = 20 + 1 = 21 $.

Therefore, we can write this as:

$\frac{1}{10} \beta \left( \frac{1}{10}, 21 \right)$

Comparing this to $ a \times \beta(b, c) $, we have $ a = \frac{1}{10} $, $ b = \frac{1}{10} $, and $ c = 21 $.

Now we calculate $ 100(a + b + c) $:

$100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{5} + 21 \right) = 100 \left( \frac{1}{5} + \frac{105}{5} \right) = 100 \left( \frac{106}{5} \right) = 100 \times 21.2 = 2120$

So, the answer is Option D, 2120.

$\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$

$\begin{aligned} & \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\ & \begin{array}{l} 3 A-5 B=1 \\ 5 A+3 B=0 \end{array}>A=\frac{3}{34} \quad B=\frac{-5}{34} \end{aligned}$

$\begin{aligned} & \int_0^{\pi / 4} \frac{136\left[\frac{3}{34}(3 \sin x+5 \cos x)-\frac{5}{34}(3 \cos x-5 \sin x)\right]}{3 \sin x+5 \cos x} d x \\ & \int_0^{\pi / 4} 12 d x-20 \int_0^{\pi / 4} \frac{3 \cos x-5 \sin x}{3 \sin x+5 \cos x} d x \\ & 12 \times \frac{\pi}{4}-20\left[\ln \left|\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right|-\ln 5\right] \\ & 3 \pi-20 \ln 2^{5 / 2}+20 \ln 5 \\ & \Rightarrow 3 \pi-50 \ln 2+20 \ln 5 \end{aligned}$

$\begin{aligned} & I=\int_\limits{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y \\ & =\int_\limits0^\pi\left(\frac{2 y(1+\sin y)}{1+\cos ^2 y}+\frac{-2 y(1-\sin y)}{1+\cos ^2 y}\right) d y \\ & =\int_\limits0^\pi\left(\frac{2 y+2 y \sin y-2 y+2 y \sin y}{1+\cos ^2 y}\right) d y \end{aligned}$

$I=4 \int_0^\pi\left(\frac{y \sin }{1+\cos ^2 y}\right) d y \quad \text{... (1)}$

$\begin{aligned} & I=4 \int_\limits0^\pi\left(\frac{(\pi-y) \sin (\pi-)}{1+\cos ^2(\pi-y)}\right) d y \\ & I=4\left\lfloor\int_\limits0\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y-\int_\limits0^\pi \frac{y \sin y}{1+\cos ^2 y} d y\right\rfloor \quad \text{... (2)} \end{aligned}$

Adding equation (1) and (2)

$\begin{aligned} & 2 I=4 \int_\limits0^\pi\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y \\ & I=2 \pi \int_\limits0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\ & =2 \pi \times \frac{\pi}{2} \\ & =\pi^2 \end{aligned}$

Given $f(x)=\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t$

Now, $\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}\left(\frac{0}{0} \text { form }\right)$

$\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t}{x^3} \\ & =\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}\left(\frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0} \frac{1+\cos \left(1-e^x\right)\left(-e^x\right)}{6 x}\left(\frac{0}{0}\right) \\ & =\lim _{x \rightarrow 0} \frac{-\sin \left(1-e^x\right)\left(e^x\right)^2+\cos \left(1-e^x\right)\left(-e^x\right)}{6} \\ & =-\frac{1}{6} \end{aligned}$

$\begin{aligned} & \text { Given, } \int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x=\frac{2}{\pi} \\ & \begin{aligned} I & =\int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x \\ \Rightarrow I & =\int_\limits0^1\left(\frac{\cos \alpha x}{1+3^x}+\frac{\cos \alpha x}{1+3^{-x}}\right) d x \\ & =\int_\limits0^1 \cos \alpha x d x \\ & =\left(\frac{\sin \alpha x}{\alpha}\right)_0^1 \\ & =\frac{\sin \alpha}{\alpha} \\ \Rightarrow & \frac{\sin \alpha}{\alpha}=\frac{2}{\pi} \\ \Rightarrow & \alpha=\frac{\pi}{2} \end{aligned} \end{aligned}$

$f(x)=\left\{\begin{array}{cc} -2 & -2 \leq x \leq 0 \\ x-2 & 0< x \leq 2 \end{array} h(x)=f|x|+|f(x)|\right.$

$\begin{aligned} & h(x)=\left\{\begin{array}{cc} -x-2+2=-x & -2 \leq x \leq 0 \\ 0 & 0< x \leq 2 \end{array}\right. \\ & \therefore \int_\limits{-2}^2 h(x) d x=\int_\limits{-2}^0-x d x+\int_\limits0^2 0 d x \\ & \left.\frac{x^2}{2}\right|_{-2} ^0=\frac{4}{2}=2 \end{aligned}$