Definite Integration

Given,

$\int_1^e {{{{{\left( {\log _e^x} \right)}^{1/2}}} \over {x{{\left( {a - {{\left( {\log _e^x} \right)}^{3/2}}} \right)}^2}}} = 1} $

Let $\log _e^x = t$

$ \Rightarrow x = {e^t}$

$ \Rightarrow dx = {e^t}dt$

When $x = 1$ then $t = \log _e^1 = 0$

and when $x = e$ then $t = \log _e^e = 1$

$ \Rightarrow \int_0^1 {{{\sqrt t \,.\,{e^t}dt} \over {{e^t}{{(a - t\sqrt t )}^2}}} = 1} $

$ \Rightarrow \int_0^1 {{{\sqrt t \,dt} \over {{{(a - t\sqrt t )}^2}}} = 1} $

Let ${t^{1/2}} = z$

$ \Rightarrow {1 \over {2\sqrt t }}dt = dz$

$ \Rightarrow dt = 2\sqrt t \,dz \Rightarrow dt = 2z\,dz$

When $t = 0$, then $z = 0$

When $t = 1$, then $z = {1^{1/2}} = 1$

$\therefore$ $\int_0^1 {{{z \times 2z\,dz} \over {{{(a - {z^3})}^2}}} = 1} $

$ \Rightarrow \int_0^1 {{{2{z^2}\,dz} \over {{{(a - {z^3})}^2}}} = 1} $

Now let $a - {z^3} = p$

$ \Rightarrow - 3{z^2}\,dz = dp$

When $z = 0$ then $p = a - 0 = a$

When $z = 1$ then $p = a - {1^3} = a - 1$

$\therefore$ $\int_a^{a - 1} {{{2\,.\,\left( { - {{dp} \over 3}} \right)} \over {{p^2}}} = 1} $

$ \Rightarrow - {2 \over 3}\int_a^{a - 1} {{{dp} \over {{p^2}}} = 1} $

$ \Rightarrow - {2 \over 3}\left[ { - {1 \over p}} \right]_a^{a - 1} = 1$

$ \Rightarrow - {2 \over 3}\left[ { - {1 \over {a - 1}} + {1 \over a}} \right] = 1$

$ \Rightarrow - {2 \over 3}\left[ {{{ - a + a - 1} \over {a(a - 1)}}} \right] = 1$

$ \Rightarrow {2 \over 3}\left( {{1 \over {a(a - 1)}}} \right) = 1$

$ \Rightarrow 3{a^2} - 3a = 2$

$ \Rightarrow 3{a^2} - 3a - 2 = 0$

$ \Rightarrow a = {{3\, \pm \,\sqrt {9 - 4(3)( - 2)} } \over {2\,.\,3}}$

$\therefore$ $a = {{3\, \pm \,\sqrt {33} } \over 6},\,{{3 - \sqrt {33} } \over 6}$

$ \text { } \begin{aligned} & \text { Let } I=\int_0^4 \| x-2|-x| d x \\ & I=\int_0^2|-(x-2)-x| d x+\int_2^4|x-2-x| d x \\ &=\int_0^2|2-2 x| d x+\int_2^4 2 d x \\ &=\int_0^1(2-2 x) d x+\int_1^2-(2-2 x) d x+\int_2^4 2 d x \\ &=\left[2 x-x^2\right]_0^1-\left[2 x-x^2\right]_1^2+2[x]_2^4 \\ &=[2-1-0]-[4-4-2+1]+2[4-2] \\ &=1+1+4 \\ &=6 \end{aligned} $

$\int_{-a}^a f(x) \cdot d x=\int_0^a f(x) d x+\int_0^a g(x) d x$

As we know,

$ \int_{-a}^a f(x) d x=\int_{-a}^0 f(x) d x+\int_0^a f(x) d x $

In 2nd term, on putting $x=-t$

$ \Rightarrow \quad d x=-d t $

Limits when $x=0, t=0$

When $x=-a, t=a$

$ \begin{aligned} \therefore \quad \int_{-a}^a f(x) d x & =\int_a^0 f(-t),(-d t)+\int_0^a f(x) d x \\ & =\int_0^a f(x) d x-\int_a^0 f(-t) d t \\ & =\int_0^a f(x)+\int_0^a f(-t) d t \\ & =\int_0^a f(x)+\int_0^a f(-x) d x \\ \therefore g(x)=f(-x) . & \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x \\ & f: R \rightarrow R, f(x)=x^2+2 \\ & \text { Now, } \lim _{n \rightarrow \infty} \frac{3}{n}\left[\begin{array}{r} f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right) \\ +\ldots+f(7) \end{array}\right] \\ & =\lim _{n \rightarrow \infty} \frac{3}{n}\left[\begin{array}{r} \left(\frac{7}{n}\right)^2+2+\left(\frac{14}{n}\right)^2+2+\left(\frac{21}{n}\right)^2 \\ +2+\ldots+\left(\frac{7 n}{n}\right)^2+2 \end{array}\right] \\ & =\lim _{n \rightarrow \infty} \frac{3}{n}\left[\begin{array}{r} 2 n+\left(\frac{7}{n}\right)^2\left(1^2+2^2+3^2\right. \\ +\ldots+n^2 \end{array}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{n \rightarrow \infty} \frac{3}{n}\left[2 n+\frac{49}{n^2} \frac{n(n+1)(2 n+1)}{6}\right] \\ & =\lim _{n \rightarrow \infty} 3\left[2+\frac{49}{6} \frac{n^2\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{n^2}\right] \\ & =3\left[2+\frac{49}{6} \times 2\right] \\ & =55 \end{aligned} $

Given,

$ \begin{aligned} & f(x)=\left|\begin{array}{ccc} 2 \cos ^2 x & \sin 2 x & \sin x \\ \sin 2 x & 2 \sin ^2 x & -\cos x \\ \sin x & -\cos x & 0 \end{array}\right| \\ & \text { Applying } R_2 \rightarrow R_2-2 \cos x R_3 \\ & f(x)=\left|\begin{array}{ccc} 2 \cos ^2 x & \sin 2 x & \sin x \\ 0 & 2 & -\cos x \\ \sin x & -\cos x & 0 \end{array}\right| \end{aligned} $

Expanding along $R_2$,

$ \begin{aligned} & f(x)=\left[0+2\left(0-\sin ^2 x\right)\right. \\ &+ \cos x\left(-2 \cos ^3 x-\sin x \sin 2 x\right] \\ & \Rightarrow f(x)=-2 \sin ^2 x \\ &-2 \cos ^2 x\left(\sin ^2 x+\cos ^2 x\right)=-2 \\ & \Rightarrow f^{\prime}(x)=0 \\ & \therefore \int_0^{\pi / 4}\left[2|f(x)|+5 f^{\prime}(x)\right] d x \\ &=\int_0^{\pi / 4}(4+0) d x=\pi \end{aligned} $

$ \text { } \begin{aligned}Let \,\,\, I & =\int_0^3\left(\sin \left(\frac{\pi}{3} x\right)-\cos \left(\frac{\pi}{3} x\right)\right) d x \\ \quad I & =\left[\frac{-\cos \frac{\pi}{3} x}{\frac{\pi}{3}}-\frac{\sin \left(\frac{\pi}{3} x\right)}{\frac{\pi}{3}}\right]_0^3 \end{aligned} $

$ \begin{aligned} & =-\frac{3}{\pi}\left[\cos \frac{\pi}{3} x+\sin \left(\frac{\pi x}{3}\right)\right]_0^3 \\ & =-\frac{3}{\pi}[(\cos \pi+\sin \pi)-(\cos 0+\sin 0)] \\ & =-\frac{3}{\pi}[-1-1]=\frac{6}{\pi} \end{aligned} $

$ \begin{aligned} & \text { Let } I=\int_0^{\pi / 2} \sin ^4 \theta \cos ^3 \theta d \theta \\ & =\int_0^{\pi / 2} \sin ^4 \theta+\left(1-\sin ^2 \theta\right) \cos \theta d \theta \end{aligned} $

Let $\sin \theta=t$, then $\cos \theta d \theta=d t$

When $\theta \rightarrow 0, t \rightarrow 0$

$ \begin{aligned} & \text { and } \theta \rightarrow \frac{\pi}{2}, t \rightarrow 1 \\ & \qquad \begin{aligned} I & =\int_0^1 t^4\left(1-t^2\right) d t \\ & =\int_0^1\left(t^4-t^6\right) d t \\ & =\left[\frac{t^5}{5}-\frac{t^7}{7}\right]_0^1=\frac{1}{5}-\frac{1}{7} \\ & =\frac{2}{35} \end{aligned} \end{aligned} $

Let $1+x^2=t$

$2 x d x=d t$

When $x=0$ and $t=1$, when $x=1, t=2$

$ \begin{aligned} \therefore \exp \left[\int_1^2 \log t\right] & =\exp \left[[t \log t-t]_1^2\right] \\ & =\exp [2 \log 2-2+1] \\ & =\exp [\log 4-1] \\ & =e^{\log 4} \times e^{-1}=\frac{4}{e} \end{aligned} $

$ \text { } \begin{aligned}Let\,\, I & =\int_0^{2 a} f(x) d x \\ & =\int_0^a f(x) d x+\int_a^{2 a} f(x) d x \ldots \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned}Let\,\, I_1 & =\int_a^{2 a} f(x) d x \\ Let\,\,\,x & =t+a \\ d x & =d t \end{aligned} \end{aligned} $

$\,\,\,\,\,\,\,\,\,\, I_1=\int_0^a f(t+a) d t=\int_0^a f(a+x) d x $

$ \therefore \quad I=\int_0^a f(x) d x+\int_0^a f(a+x) d x $

Let $I=\int_1^2 x \sqrt{4-x^2} d x$

On putting $4-x^2=1$,

$ \begin{aligned} & -2 x d x =d t \\ \Rightarrow &\,\,\,\, x d x =-\frac{d t}{2} \end{aligned} $

When $x \rightarrow 1, t \rightarrow 3$

$ \begin{aligned} &\\ &\begin{aligned} \text { } k= & \int_{\frac{-3}{2}}^{\frac{3}{2}}[2 x-3] d x=\int_{\frac{-3}{2}}^{\frac{3}{2}}([2 x]-3) d x \\ = & \int_{-3 / 2}^{-1}(-3) d x+\int_{-1}^{-1 / 2}(-2) d x+\int_{\frac{-1}{2}}^0(-1) d x+\int_0^{1 / 2}(0) d x +\int_{\frac{1}{2}}^1 d x+\int_1^{3 / 2} 2 d x-\int_{\frac{-3}{2}}^{3 / 2} 3 d x \\ = & \frac{-3}{2}-1-\frac{1}{2}+\frac{1}{2}+1-9=\frac{-3}{2}-9 \\ \therefore & \left|k+\frac{1}{2}\right|=\left|-\frac{3}{2}-9+\frac{1}{2}\right|=10 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { Given, } \int_1^4\left(x+\sqrt{x}+\frac{1}{x}\right) d x-\int_1^{2 \log 2} 1 d x \\ & =\left[\frac{x^2}{2}+\frac{2}{3} x^{\frac{3}{2}}+\ln x\right]_1^4-[x]_1^{2 \log 2} \\ & =\left(8+\frac{16}{3}+\ln 4\right)-\left(\frac{1}{2}+\frac{2}{3}+0\right) \\ & -(2 \log 2-1)=\frac{17}{2}+\frac{14}{3}=\frac{79}{6} \end{aligned} $

Let

$ I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x $

Let

$ \begin{aligned} f(x) & =\frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) \\ f(-x) & =\frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4-\sin x}{4+\sin x}\right)\right) \\ & =\frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}-\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) \end{aligned} $

$ \begin{aligned} f(x) & +f(-x)=\frac{6}{\pi(2-\cos 2 x)} \\ I & =\int_{-\pi / 4}^{\pi / 4} f(x) d x \\ & =\int_0^{\pi / 4} f(x)+f(-x) d x \\ & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{1}{2-\cos 2 x} d x \\ I & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{1}{2-\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)} d x \\ & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{1+\tan ^2 x}{1+3 \tan ^2 x} d x \\ & =\frac{6}{\pi} \int_0^{\pi / 4} \frac{\sec ^2 x}{1+3 \tan ^2 x} d x \end{aligned} $

$ \begin{aligned} &\text { On putting } \tan x=t\\ &\begin{aligned} & \sec ^2 x d x=d t \\ & I=\frac{6}{\pi} \int_0^1 \frac{d t}{1+(\sqrt{3} t)^2}=\frac{6}{\pi} \cdot \frac{1}{\sqrt{3}}\left[\tan ^{-1} \sqrt{3} t\right]_0^1 \\ & =\frac{2 \sqrt{3}}{\pi}\left[\tan ^{-1} \sqrt{3}-0\right] \\ & \quad I=\frac{2 \sqrt{3}}{\pi} \times \frac{\pi}{3}=\frac{2}{\sqrt{3}} \\ & \Rightarrow \quad I^2=\frac{4}{3} \\ & \quad 3 I^2=4 \end{aligned} \end{aligned} $

Given, $I=\int_\limits0^T f(x) d x$

If $f(x+T)=f(x)$

Now, $\int_\limits0^{5 T} f(2 x) d x$

On putting $2 x=y$

$\Rightarrow \quad d x=\frac{1}{2} d y$

$\frac{1}{2} \int_\limits0^{10 T} f(y) d y=\frac{10 I}{2}=5 I$

$\begin{aligned} \text { Let } I & =\int_\limits1^3 x \sqrt[n]{x^2-1} d x=6 \\ & =\frac{1}{2} \int_\limits1^3 2 x\left(x^2-1\right)^{\frac{1}{n}} d x=6 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad\left[\frac{\left(n^2-1\right)^{\frac{1}{n}+1}}{\frac{1}{n}+1}\right]_1^3=12 \\ & \Rightarrow \quad \frac{(8)^{\frac{1}{n}+1}}{\frac{1}{n}+1}=12 \\ & \Rightarrow \quad \frac{(8)^{\frac{1}{n}}}{(n+1)}(n)=\frac{3}{2} \end{aligned}$

Now, on putting $n=3$, L.H.S = R. H. S

$\therefore n=3$

$\begin{aligned} & \int_\limits{-1}^1(x[1+\sin \pi x]+1) d x \\ & I=\int_\limits{-1}^1(x[1+\sin \pi x]+1) d x \end{aligned}$

We know that, $-1 \leq \sin \pi x \leq 1$

$\begin{aligned} & I=\int_\limits{-1}^0(x[1+\sin \pi x]+1) d x+\int_\limits0^1(x(1+\sin \pi x)+1) d x \\ & \Rightarrow I=\int_\limits{-1}^0(x \cdot 0+1) d x+\int_\limits0^1(x+1) d x \\ & \Rightarrow \quad I=[x]_{-1}^0+\left[\frac{x^2}{2}+x\right]_0^1 \\ & \Rightarrow \quad I=0-(-1)+\left[\frac{1}{2}+1\right] \\ & \Rightarrow \quad I=2+\frac{1}{2}=\frac{5}{2} \end{aligned}$

$\text { Here, } \lim\limits_{n \rightarrow \infty}\left[\begin{array}{cc} \frac{n}{(n+1) \sqrt{2 n+1}}+\frac{n}{(n+2) \sqrt{2(n+2)}} \\ & +\ldots \text { upto } n \text { terms } \end{array}\right]$

$ \begin{aligned} & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{n^2}{(n+k) \sqrt{k(2 n+k)}} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{n^2}{(n+k) \sqrt{k(2 n+k)}} \times \frac{n^2}{n^2} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\left(\frac{n+k}{n}\right) \sqrt{\frac{k}{n}\left(\frac{2 n+k}{n}\right)}} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\left(1+\frac{k}{n}\right) \sqrt{\frac{k}{n}\left(2+\frac{k}{n}\right)}} \\ & =\int_\limits0^1 \frac{1}{(1+x) \sqrt{x(2-x)}} d x=\int_\limits0^1 f(x) d x \\ f(x) & =\frac{1}{(1+x) \sqrt{2 x+x^2}} \end{aligned}$

$\begin{aligned} I_n & =\int_0^{\pi / 4} \tan ^n x d x \\ I_{n+2} & =\int_0^{\frac{\pi}{4}} \tan ^{n+2} x d x \end{aligned}$

$\begin{array}{r} \text { Now, } I_n+I_{n+2}=\int_0^{\frac{\pi}{4}} \tan ^n x d x+\int_0^{\frac{\pi}{4}} \tan ^n x \tan ^2 x d x \\ =\int_0^{\frac{\pi}{4}} \tan ^n x\left(1+\tan ^2 x\right) d x=\int_0^{\frac{\pi}{4}} \tan ^n x \sec ^2 x d x \end{array}$

$\begin{aligned} & \text { Put } \tan x=t \Rightarrow \sec ^2 x d x=d t \\ & \begin{aligned} & \therefore I_n+I_{n+2}=\int_0^1 t^n d t=\frac{1}{n+1} \\ & \text { Now, } \frac{1}{I_2+I_4}+\frac{1}{I_3+I_5}+\frac{1}{I_4+I_6} \\ &=\frac{1}{1 / 3}+\frac{1}{1 / 4}+\frac{1}{1 / 5} \\ &=3+4+5=12=\frac{1}{I_{11}+I_{13}} \end{aligned} \end{aligned}$

$\begin{aligned} & \text { Let } I=\int e^{\tan ^2 \theta} \sin ^2 \theta \tan \theta d \theta \\ & =\int e^{\tan ^2 \theta} \times \sin ^2 \theta \frac{\tan \theta \sec ^2 \theta}{\left(1+\tan ^2 \theta\right)} d \theta \\ & =\frac{1}{2} \int 2 e^{\tan ^2 \theta} \times \frac{\tan ^2 \theta}{\left(1+\tan ^2 \theta\right)} \times \frac{\tan \theta \sec ^2 \theta}{\left(1+\tan ^2 \theta\right)} d \theta \end{aligned}$

Let $\tan ^2 \theta=t$

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

When $\theta=0$, then $t=0$

and when $\theta=\frac{\pi}{4}$, then $t=1$

$\begin{aligned} \therefore I & =\frac{1}{2} \int_0^1 e^t \times \frac{t}{(1+t)^2} d t=\frac{1}{2} \int_0^1 e^t\left[\frac{1+t-1}{(1+t)^2}\right] d t \\ & =\frac{1}{2} \int_0^1 e^t\left[\frac{1}{1+t}+\left(\frac{-1}{(1+t)^2}\right)\right] d t \end{aligned}$

$\begin{aligned} & =\frac{1}{2}\left[\frac{e^t}{1+t}\right]_0^1\left[\because \int e^x\left(f(x)+f^{\prime}(x) d x\right)=e^x f(x)+C\right] \\ & =\frac{1}{2}\left[\frac{e}{2}-1\right] \end{aligned}$

$\int_{\pi / 4}^{5 \pi / 4}(|\cos t| \sin t+|\sin t| \cos t) d t$

$\begin{aligned} & =\int_{\pi / 4}^{5 \pi / 4}|\cos t| \sin t+\int_{\pi / 4}^{5 \pi / 4}|\sin t| \cos t d t \\ & =\frac{2}{2} \int_{\pi / 4}^{\pi / 2} \cos \sin t d t+\frac{2}{2} \int_{\pi / 2}^{5 \pi / 4}(-\cos t) \sin t d t \\ & \quad+\frac{2}{2} \int_{\pi / 4}^\pi \sin t \cos t d t+\frac{2}{2} \int_\pi^{5 \pi / 4}(-\sin t) \cos t d t \\ & =\frac{1}{2} \int_{\pi / 4}^{\pi / 2} \sin 2 t d t-\frac{1}{2} \int_{\pi / 2}^{5 \pi / 4} \sin 2 t d t+\frac{1}{2} \int_{\pi / 4}^\pi \sin 2 t d t \\ & -\frac{1}{2} \int_\pi^{5 \pi / 4} \sin 2 t d t \end{aligned}$

$\begin{array}{r} =\frac{1}{2}\left[\frac{-\cos 2 t}{2}\right]_{\pi / 4}^{\pi / 2}-\frac{1}{2}\left[\frac{-\cos 2 t}{2}\right]_{\pi / 2}^{5 \pi / 4}+\frac{1}{2}\left[\frac{-\cos 2 t}{2}\right]_{\pi / 4}^\pi \\ -\frac{1}{2}\left[\frac{-\cos 2 t}{2}\right]_\pi^{5 \pi / 4} \end{array}$

$\begin{array}{r} =\frac{-1}{4}\left(\cos \pi-\frac{\cos \pi}{2}\right)+\frac{1}{4}\left(\cos \frac{5 \pi}{2}-\cos \pi\right) \\ \frac{-1}{4}\left(\cos 2 \pi-\frac{\cos \pi}{2}\right)+\frac{1}{4}\left(\cos \frac{5 \pi}{2}-\cos 2 \pi\right) . \end{array}$

$\begin{aligned} & =\frac{-1}{4}(-1-0)+\frac{1}{4}(0-(-1))-\frac{1}{4}(1-0)+\frac{1}{4}(0-1) \\ & =\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}=0 \end{aligned}$

Given, $f(x)=\max \{\sin x, \cos x\}$ and $g(x)=\min \{\sin x, \cos x\}$

$\begin{aligned} \therefore & =\int_0^\pi f(x) d x+\int_0^\pi g(x) d x=\int_0^\pi \sin d x+\int_0^\pi \cos x d x \\ & =-[\cos x]_0^\pi+[\sin x]_0^\pi=-(-1-1)+0=2 \end{aligned}$

If we consider the option (a), we have

$\begin{aligned} \lim _\limits{n \rightarrow \infty} & \frac{a^k\left(1+2^k+3^k+\ldots .+n^k\right)}{n^k+1} \\ & =\lim _\limits{n \rightarrow \infty} \frac{a^k\left(1^k+2^k+3^k+\ldots . .+n^k\right)}{n^k \cdot n} \\ & =\lim _\limits{n \rightarrow \infty} \sum_{r=1}^n \frac{a^k r^k}{n^k} \cdot \frac{1}{n} \\ & =\lim _\limits{n \rightarrow \infty} \sum_{r=1}^n a^k\left(\frac{r}{n}\right)^k \cdot \frac{1}{n} \end{aligned}$

On putting $\frac{1}{n} \rightarrow d x, \frac{r}{n} \rightarrow x$ and $\Sigma \rightarrow \int$

$=\int_0^1 a^k \cdot x^k \cdot d x$

$\begin{aligned} & f(x)=x^2, g(x)=\cos x \\ & 18 x^2-9 \pi x+\pi^2=0 \\ & \Rightarrow \quad x=\frac{9 \pi \pm \sqrt{81 \pi^2-72 \pi^2}}{36} \\ & \Rightarrow \quad x=\frac{9 \pi \pm 3 \pi}{36} \\ & \quad x=\frac{12 \pi}{36} \\ & \Rightarrow \quad \beta=\frac{\pi}{3} \end{aligned}$

and $x=\frac{6 \pi}{36} \Rightarrow \alpha=\frac{\pi}{6} \quad [\because \alpha<\beta]$

Now, $g \circ f(x)=g\left(f(x)=g\left(x^2\right)=\cos x^2\right.$

$\begin{aligned} & \therefore \quad \int_\limits\alpha^\beta x(g \circ f(x)) d x=\int_\limits{\pi / 6}^{\pi / 3} x \cdot \cos x^2 d x \\ &=\frac{1}{2} \int_\limits{\pi / 6}^{\pi / 3} \cos \left(x^2\right) \cdot 2 x d x \\ &=\frac{1}{2}\left[\sin x^2\right]_{\pi / 6}^{\pi / 3} \\ &=\frac{1}{2}\left[\sin \frac{\pi^2}{9}-\sin \frac{\pi^2}{36}\right] \end{aligned}$

$\text { Let } I=\int_0^\pi x\left(\sin ^2(\sin x)+\cos ^2(\cos x)\right) d x \quad \text{....(i)}$

$I=\int_0^\pi(\pi-x)\left(\sin ^2(\sin x)+\cos ^2(\cos x)\right) d x\quad \text{... (ii)}$

Adding Eqs. (i) and (ii), we get

$\begin{aligned} 2 I & =\pi \int_0^\pi\left(\sin ^2(\sin x)+\cos ^2(\cos x)\right) d x \\ 2 I & =2 \pi \int_0^{\pi / 2}\left(\sin ^2(\sin x)+\cos ^2(\cos x)\right) d x \\ I & =\pi \int_0^{\pi / 2} \sin ^2(\sin x)+\int_0^{\pi / 2} \cos ^2(\cos x) d x \\ I & =\pi \int_0^{\pi / 2} \sin ^2(\cos x)+\int_0^{\pi / 2} \cos ^2(\cos x) d x \\ & =\pi \int_0^{\pi / 2} 1 d x \end{aligned}$

$\begin{aligned} & I=\pi \times \frac{\pi}{2} \\ & I=\frac{\pi^2}{2} \end{aligned}$

$I = {\pi ^2}\int_0^2 {\sin \left( {{{\pi x} \over 2}} \right){{(x - [x])}^{[x]}}dx} $

$ = {\pi ^2}\int_0^1 {\sin \left( {{{\pi x} \over 2}} \right){x^0}dx + {\pi ^2}\int_1^2 {\sin \left( {{{\pi x} \over 2}} \right)(x - 1)dx} } $

$ = {\pi ^2}\left[ {{{ - 2} \over \pi }\cos {{\pi x} \over 2}} \right]_0^1 + {\pi ^2}\left[ {(x - 1){2 \over \pi }\left( { - \cos {{\pi x} \over 2}} \right)} \right]_1^2 + {\pi ^2}\int_1^2 {{2 \over \pi }\cos {{\pi x} \over 2}dx} $

$ = \left. {{\pi ^2}\left( {{2 \over \pi }} \right) + {{2{\pi ^2}} \over \pi }(1 - 0) + 2\pi \,.\,{2 \over \pi }\left( {\sin {{\pi x} \over 2}} \right)} \right|_1^2$

$ = 2\pi + 2\pi + 4(0 - 1) = 4\pi - 4 = 4(\pi - 1)$

$I = \int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta } $

$I = \int\limits_0^1 {{x \over {{e^x}}}dx + \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx + \int\limits_2^3 {{{x + 2} \over {{e^{x - 2}}}}dx + \int\limits_3^4 {{{x + 3} \over {{e^{x - 3}}}}dx + \int\limits_4^5 {{{x + 4} \over {{e^{x - 4}}}}dx} } } } } $

Let $I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5}$

Here, ${I_2} = \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx} $ Put $x = t + 1 \Rightarrow dx = dt$

$ = \int\limits_0^1 {{{t + 2} \over {{e^t}}}dt = \int\limits_0^1 {{t \over {{e^t}}}dt + \int\limits_0^1 {{2 \over {{e^t}}}dt} } } $

${I_2} = {I_1} + 2\int\limits_0^1 {{e^{ - t}}dt = {I_1} + 2(1 - {e^{ - 1}})} $

Similarly,

${I_3} = {I_1} + 4(1 - {e^{ - 1}})$

${I_4} = {I_1} + 6(1 - {e^{ - 1}})$

${I_5} = {I_1} + 8(1 - {e^{ - 1}})$

$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5} = 5{I_1} + (2 + 4 + 6 + 8)(1 - {e^{ - 1}})$

$ = 5{I_1} + 20(1 - {e^{ - 1}})$

${I_1} = \int_0^1 {x{e^{ - 1}}dx = - [{e^{ - x}}(x + 1)_0^1 = 1 - 2{e^{ - 1}}} $

$\therefore$ $5{I_1} + 20(1 - {e^{ - 1}}) = 5(1 - 2{e^{ - 1}}) + 20(1 - {e^{ - 1}}) = 25 - 30{e^{ - 1}}$

$\therefore$ $\alpha$ = $-$30, $\beta$ = 25

Also it satisfy $5\alpha + 6\beta = 0$

Now, ${(\alpha + \beta )^2} = {( - 30 + 25)^2} = {( - 5)^2} = 25$

$\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} $

$ = \sqrt {{{\left( {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right)}^2}} $

$ = \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|$

$ = \left| {{{{{(x + 1)}^2} - {{(x - 1)}^2}} \over {{x^2} - 1}}} \right|$

$ = \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|$

$ = \left| {{{4x} \over {{x^2} - 1}}} \right|$

$\therefore$ $I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx} $

$I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $

Let $f(x) = \left| {{{4x} \over {{x^2} - 1}}} \right|$

$\therefore$ $f( - x) = \left| {{{4( - x)} \over {{{( - x)}^2} - 1}}} \right| = \left| {{{ - 4x} \over {{x^2} - 1}}} \right| = \left| {{{4x} \over {{x^2} - 1}}} \right|$

$ \Rightarrow f(x) = f( - x)$

$\therefore$ f(x) is a even function.

$\therefore$ $I = 2\int_0^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $

Using property, If f(x) is an even function then,

$\int_{ - a}^a {f(x) = 2\int_0^a {f(x)dx} } $

x > 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$ x² > 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$ x² $-$ 1 < 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$ \Rightarrow {1 \over {{x^2} - 1}} < 0$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$ \Rightarrow {{4x} \over {{x^2} - 1}} < 0$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\therefore$ $\left| {{{4x} \over {{x^2} - 1}}} \right| = - {{4x} \over {{x^2} - 1}}$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$

$\therefore$ $I = 2\int_0^{{1 \over {\sqrt 2 }}} { - \left( {{{4x} \over {{x^2} - 1}}} \right)dx} $

$ = - 4\int_0^{{1 \over {\sqrt 2 }}} {{{2x} \over {{x^2} - 1}}dx} $

$ = - 4\left[ {{{\log }_e}\left| {{x^2} - 1} \right|} \right]_0^{{1 \over {\sqrt 2 }}}$

$ = - 4\left[ {\log \left| {{1 \over 2} - 1} \right| - \log \left| { - 1} \right|} \right]$

$ = - 4{\log _e}\left| { - {1 \over 2}} \right|$

$ = - 4{\log _e}{1 \over 2}$

$ = - 4\log _e^{{2^{ - 1}}}$

$ = 4\log _e^2$

$ = \log _e^{{2^4}}$

$ = \log _e^{16}$