Definite Integration

$\int\limits_{-1 / 24}^{1 / 24} \sec x \log \left(\frac{1-x}{1+x}\right) d x$

Here, $f(x)=\sec x \log \left(\frac{1-x}{1+x}\right)$

Now, $f(-x)=\sec (-x) \log \left(\frac{1-(-x)}{1-x}\right)$

$ \begin{aligned} & =\sec x \log \left(\frac{1+x}{1-x}\right) \\ & =-\sec x \log \left(\frac{1-x}{1+x}\right)=-f(x) \end{aligned} $

Thus, the given function is an odd function

$\therefore \quad \int\limits_{\frac{-1}{24}}^{\frac{1}{24}} \sec x \log \left(\frac{1-x}{1+x}\right) d x=0$

$\left[\because \int\limits_{-a}^a f(x) d x=0\right.$, if $f(x)$ is an odd function]

$I=\int_0^5[x] d x$

$ =\int_0^1 0 d x+\int_1^2 1 d x+\int_2^3 2 d x+\int_3^4 3 d x+\int_4^5 4 d x $

$=0+(2-1)+2(3-2)+3(4-3)+4(5-4)$ $=1+2+3+4=10$

$I=\int_0^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x\quad \ldots \text { (i) } $

$ \begin{aligned} & =\int_0^{\pi / 2} \frac{1}{\sqrt{1+\tan \left(\frac{\pi}{2}-x\right)}} d x \\ & =\int_0^{\pi / 2} \frac{1}{\sqrt{1+\cot x}} d x \\ & =\int_0^{\pi / 2} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x \quad \ldots \text { (ii) } \end{aligned} $

$\therefore$ From Eqs. (i) and (ii), we get

$ 2 I=\int_0^{\pi / 2} 1 d x=[x]_0^{\pi / 2}=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4} $

To solve the integral $ I = \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} \, dx $, we can use symmetry properties of definite integrals. According to the property:

$ \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx $

Apply this property to transform the integral:

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

Adding both expressions for $ I $, we have:

$ 2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} \, dx $

Now, perform substitution to evaluate the new integral. Let $\cos x = t$, then $ -\sin x \, dx = dt $. The limits of integration change as $ x $ changes from 0 to $\pi$, causing $ t $ to vary from 1 to -1. Thus, we have:

$ -\pi \int_1^{-1} \frac{dt}{1+t^2} $

Reversing the limits, the expression becomes:

$ \pi \int_{-1}^{1} \frac{dt}{1+t^2} $

This integral evaluates to:

$ \pi \left[ \tan^{-1} t \right]_{-1}^{1} $

Simplifying:

$ = \pi \left( \tan^{-1}(1) - \tan^{-1}(-1) \right) $

Since $\tan^{-1}(1) = \frac{\pi}{4}$, it follows that:

$ = \pi \left( \frac{\pi}{4} + \frac{\pi}{4} \right) = \pi \times \frac{\pi}{2} = \frac{\pi^2}{2} $

Thus, the original integral $ I $ is:

$ I = \frac{1}{2} \times \frac{\pi^2}{2} = \frac{\pi^2}{4} $

We have $I=\int_{-\pi}^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x$

Let $f(x)=\frac{x \sin ^3 x}{4-\cos ^2 x}$

$ f(-x)=\frac{-x \times \sin ^3(-x)}{4-\cos ^2(-x)}=\frac{x \sin ^3 x}{4-\cos ^2 x}=f(x) $

$\therefore f(x)$ is an even function

$ \begin{aligned} & I=2 \int_0^\pi \frac{x \sin ^3 x}{4-\cos ^2 x} d x \\ & I=2 \int_0^\pi \frac{(\pi-x) \sin ^3 x}{4-\cos ^2 x} d x \end{aligned} $

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

$ \begin{aligned} & 2 I=2 \pi \int_0^\pi \frac{\sin ^3 x}{4-\cos ^2 x} d x \\ & I=\pi \int_0^\pi \frac{\sin ^3 x}{4-\cos ^2 x} d x \end{aligned} $

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

When $x \rightarrow 0 t \rightarrow 1$ and $x \rightarrow \pi, t \rightarrow-1$

$ \begin{aligned} I & =-\pi \int_1^{-1} \frac{\left(1-t^2\right)}{4-t^2} d t \\ & =\pi \int_{-1}^1\left(\frac{-3}{4-t^2}+1\right) d t \\ & =\pi\left[\frac{-3}{4} \log \left|\frac{2+t}{2-t}\right|+t\right]_{-1}^1 \\ & =\pi\left[\left(\frac{-3}{4} \log 3+1\right)-\left(-\frac{3}{4} \log \frac{1}{3}-1\right)\right] \\ & =\pi\left[-\frac{3}{4} \log 3+1-\frac{3}{4} \log 3+1\right] \\ & =2 \pi\left[1-\frac{3}{4} \log 3\right] \end{aligned} $

To solve the integral, we have:

$ I = \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{1}{\sqrt[5]{x^{30} + x^{25}}} \, dx $

First, we can rewrite the integrand as:

$ \int_{\frac{1}{\sqrt[5]{31}}}^{\frac{1}{\sqrt[5]{242}}} \frac{1}{x^6 \sqrt[5]{1 + \frac{1}{x^5}}} \, dx $

To simplify, we perform a substitution. Let:

$ 1 + \frac{1}{x^5} = t^5 $

Differentiating both sides gives:

$ -\frac{5}{x^6} \, dx = 5t^4 \, dt $

Thus, the differential becomes:

$ dx = -x^6 \cdot \frac{1}{5t^4} \, dt $

We need to adjust the limits of integration. When $ x \rightarrow \frac{1}{\sqrt[5]{31}} $, then $ t \rightarrow 2 $. When $ x \rightarrow \frac{1}{\sqrt[5]{242}} $, then $ t \rightarrow 3 $.

The integral now becomes:

$ I = -\int_2^3 \frac{1}{t} \times t^4 \, dt $

This simplifies to:

$ I = -\int_2^3 t^3 \, dt $

$ I = -\left[ \frac{t^4}{4} \right]_2^3 $

Calculating the definite integral gives:

$ I = -\left( \frac{3^4}{4} - \frac{2^4}{4} \right) $

$ I = -\left( \frac{81}{4} - \frac{16}{4} \right) $

$ I = -\left( \frac{65}{4} \right) $

Thus, the evaluated integral is:

$ I = \frac{-65}{4} $

$ \begin{aligned} &\text {Let, }\\ &\begin{aligned} & P=\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right)\right. \\ & \left.\ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n} \\ & \log P=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\log \left(1+\frac{1}{n^2}\right)\right. \\ & \left.+\log \left(1+\frac{2^2}{n^2}\right)+\ldots \log \left(1+\frac{n^2}{n^2}\right)\right] \\ & \log P=\lim _{n \rightarrow-\infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r^2}{n^2}\right) \\ & =\int_0^1 \log \left(1+x^2\right) d x \\ & =\left[x \log \left(1+x^2\right)\right]_0^1-\int_0^1 \frac{2 x^2}{1+x^2} d x \\ & =\log 2-2 \int_0^1 d x+2 \int_0^1 \frac{x^2+1-1}{1+x^2} d x \\ & =\log 2-2 \int_0^1 d x+2 \int_0^1 \frac{d x}{1+x^2} \\ & =\log 2-z x]_0^1+2\left[\tan ^{-1} a\right]^{0^1} \\ & \log P=\log 2-2+\frac{\pi}{2} \\ & P=e^{\log 2-2+\frac{\pi}{2}}=e^{\log 2} \cdot e^{-2+\frac{\pi}{2}} \\ & P=2 e^{-2+\frac{\pi}{2}} \\ & \therefore \quad a=2 \quad b=-2+\frac{\pi}{2} \\ & a+b=2-2+\frac{\pi}{2}=\frac{\pi}{2} \end{aligned} \end{aligned} $

Let $I=\int_0^\pi x \sin ^4 x \cos ^6 x d x...(i)$

$ I=\int_0^\pi(\pi-x) \sin ^4 x \cos ^6 x d x ...(ii)$

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

$ \begin{aligned} & I=\pi \int_0^\pi \sin ^4 x \times \cos ^6 x d x \\ & I=\frac{\pi}{2} \times 2 \int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^6 x d x \\ & I=\pi \int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^6 x d x \end{aligned} $

$=\frac{\pi\left[\frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi} \times \frac{5}{2} \times \frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi}\right]}{2 \times 120}=\frac{3 \pi^2}{512}$

To determine $ I_n $, we start with the given integral:

$ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx $

Rewrite this as:

$ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \times (\sec^2 x - 1) \, dx $

Breaking it down, we have:

$ I_n = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx $

Therefore, we can express this as:

$ I_n + I_{n-2} = \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx $

When integrating, we find:

$ \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx = \left[ \frac{\tan^{n-1} x}{n-1} \right]_0^{\frac{\pi}{4}} = \frac{1}{n-1} $

Hence, we deduce:

$ I_n + I_{n-2} = \frac{1}{n-1} $

By setting $ n = 13 $, it follows that:

$ I_{13} + I_{11} = \frac{1}{12} $

Thus, the sum $ I_{13} + I_{11} $ is:

$ \frac{1}{12} $

$ \begin{aligned} & \lim _{n \rightarrow \infty} x^4\left[\begin{array}{l} \frac{1}{x^5}+\frac{1}{\left(x^2+1\right)^{5 / 2}}+\frac{1}{\left(x^2+4\right)^{5 / 2}} \\ +\frac{1}{\left(x^2+9\right)^{5 / 2}}+\ldots+\frac{1}{4 \sqrt{2} x^5} \end{array}\right] \\ & \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{x^4}{\left(x^2+r^2\right)^{5 / 2}}\left[\because \lim _{n \rightarrow \infty} \frac{x^4}{x^5}=0\right] \\ & \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left[\frac{1}{\left[1+\left(\frac{r}{n}\right)^2\right]^{5 / 2}}\right] \\ & =\int_0^1 \frac{1}{\left(1+x^2\right)^{5 / 2}} d x \\ & \text { Let } x=\tan \theta \\ & d x=\sec ^2 \theta d \theta \end{aligned} $

$\begin{aligned} x \rightarrow 0, \theta & \rightarrow 0 \text { and } \\ x \rightarrow 1, \theta & \rightarrow \frac{\pi}{4} \\ & =\int_0^{\pi / 4} \frac{1}{\left(1+\tan ^2 \theta\right)^{5 / 2}} \times \sec ^2 \theta d \theta \\ & =\int_0^{\pi / 4} \frac{1}{\sec ^3 \theta} d \theta \\ & =\int_0^{\pi / 4} \cos ^3 \theta d \theta \\ & =\frac{1}{4} \int_0^{\pi / 4}(\cos 3 \theta+3 \cos \theta) d \theta \\ & =\frac{1}{4}\left[\frac{\sin 3 \theta}{3}+3 \sin \theta\right]_0^{\pi / 4} \\ & =\frac{1}{4}\left[\frac{1}{3} \sin \left(\pi-\frac{\pi}{4}\right)+3 \sin \frac{\pi}{4}\right] \\ & =\frac{1}{4}\left[\frac{1}{3} \times \frac{1}{\sqrt{2}}+\frac{3}{\sqrt{2}}\right] \\ & =\frac{1+9}{4 \times 3 \sqrt{2}} \Rightarrow \frac{10}{4 \times 3 \sqrt{2}}=\frac{5}{6 \sqrt{2}}\end{aligned}$

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

Let $e^x=t$

$ \begin{aligned} & \Rightarrow \quad e^x d x=d t \\ & \text { When } x \rightarrow \log 4, t \rightarrow 4 \end{aligned} $

$ \begin{aligned} & x \rightarrow \log 5, t \rightarrow 5 \\ & I=\int_4^5 \frac{t+1}{(t-3)(t-2)} d t \\ & \quad=\int_4^5 \frac{4}{t-3}-\frac{3}{t-2} \end{aligned} $

[ $\because$ by partial fraction]

$\begin{aligned} & =[4 \log |t-3|-3 \log |t-2|]_4^5 \\ & =\left[\log \left\lvert\, \frac{(t-3)^4}{(t-2)^3}\right.\right]_4^5 \\ & =\log \frac{2^4}{3^3}-\log \frac{1}{2^3}=\log \frac{2^4}{3^3} \times 2^3 \\ & =\log \left(\frac{2^7}{3^3}\right)=\log \left(\frac{128}{27}\right)\end{aligned}$

$\begin{aligned} & \text { Let } I=\int_1^2 \frac{x^4-1}{x^6-1} d x \\ & \quad=\int_1^2 \frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^2-1\right)\left(x^4+x^2+1\right)} d x \\ & =\int_1^2 \frac{x^2+1}{x^4+x^2+1} d x=\int_1^2\left(\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}\right) d x \\ & =\int_1^2\left(\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-2+3}\right) d x \\ & =\int_1^2 \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+3} d x\end{aligned}$

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

When $x \rightarrow 1$, then $t \rightarrow 0$

When $x \rightarrow 2$, then $t \rightarrow \frac{3}{2}$

$ \begin{aligned} I & =\int_0^{3 / 2} \frac{1}{t^2+3} d t \\ & =\frac{1}{\sqrt{3}}\left[\tan ^{-1} \frac{1}{\sqrt{3}}\right]_0^{3 / 2} \\ & =\frac{1}{\sqrt{3}} \tan ^{-1} \frac{3}{2 \sqrt{3}} \\ & =\frac{1}{\sqrt{3}} \tan ^{-1} \frac{\sqrt{3}}{2} \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

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

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

$\int_0^\pi \log (\sin x) d x=8 k$

$\begin{aligned} & \text { Let } I=\int_0^{\pi / 4} \log (1+\tan x) d x=\int_0^{\pi / 4} \log (1+\tan \theta) d \theta \\ & =\int_0^{\pi / 4} \log (1+\tan (\pi / 4-\theta)) d \theta \\ & I=\int_0^{\pi / 4} \log \left(\frac{1+\tan \theta+1-\tan \theta}{1+\tan \theta}\right) d \theta \\ & I=\int_0^{\pi / 4} \log 2 d \theta-\int_0^{\pi / 4} \log (1+\tan \theta) d \theta \\ & I=\int_0^{\pi / 4} \log 2 d \theta-I \Rightarrow 2 I=(\log 2)\left(\frac{\pi}{4}\right) \\ & I=\frac{(\log 2) \pi}{8}=\frac{-1}{4} \times\left(\frac{-\pi}{2} \log 2\right)=-\frac{1}{8}(-\pi \log 2) \\ & =\frac{-1}{8} \times 8 k \quad\left[\begin{array}{l} \because \int_0^{\pi / 2} \log \sin x d x=\frac{-\pi}{2} \log 2 \\ =\int_0^\pi \log \sin x=-\pi \log 2 \end{array}\right] \\ & \therefore \quad I=-k \\ \end{aligned}$

$\begin{aligned} \int_0^1 x^n(1-x)^n d x & =\int_0^1(1-x)^n\left(1-(1-x)^n d x\right. \\ & =\int_0^1(1-x)^n(x)^n d x \Rightarrow k=1\end{aligned}$

$\begin{aligned} & \int_0^a \frac{d x}{4+x^2}=\frac{\pi}{8} \\ & \left.\Rightarrow \frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)\right|_0 ^a=\frac{\pi}{8} \Rightarrow \tan ^{-1}\left(\frac{a}{2}\right)=\frac{\pi}{4} \\ & \Rightarrow \quad a / 2=1 \Rightarrow a=2\end{aligned}$

$\left. {\int_1^2 {\left( {{{{x^3} - 1} \over {{x^2}}}} \right)dx = {{{x^2}} \over 2} + {1 \over x}} } \right|_1^2 = {5 \over 2} - {3 \over 2} = 1$

$\begin{aligned} & \text { } \int_0^{\pi / 2} \tan ^n x d x=k \int_0^{\pi / 2} \cot ^n(x) d x \\ & \begin{aligned} & \Rightarrow \int_0^{\pi / 2} \tan ^n\left(\frac{\pi}{2}-x\right) d x=k \int_0^{\pi / 2} \cot ^n(x) d x \\ &\left\{\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right\} \\ & \Rightarrow \int_0^{\pi / 2} \cot ^n x d x=k \int_0^{\pi / 2} \cot ^n x d x \\ & k=1\end{aligned}\end{aligned}$

$\begin{aligned} \int_0^2 x e^x d x & =\left(x e^x-e^x\right)_0^2 \\ & =\left(2 e^2-e^2\right)-\left(0 . e^0-e^0\right)=e^2+1\end{aligned}$

$\begin{aligned} & \text { Let } I=\int_2^4(|x-2|+|x-3|) d x \\ & \because x-a=\left\{\begin{array}{cc}x-a, \quad & x \geq a \\ -(x-a), \quad x < a\end{array}\right. \\ & \therefore \quad I=\int_2^4|x-2| d x+\int_2^3|x-3| d x+\int_3^4|x-3| d x \\ & =\int_2^4(x-2) d x-\int_2^3(x-3) d x+\int_3^4(x-3) d x \\ & =\left(\frac{x^2}{2}-2 x\right)_2^4-\left(\frac{x^2}{2}-3 x\right)_2^3+\left(\frac{x^2}{2}-3 x\right)_3^4 \\ & =\left[\left(\frac{16}{2}-8\right)-\left(\frac{4}{2}-4\right)\right]-\left[\left(\frac{9}{2}-9\right)-\left(\frac{4}{2}-6\right)\right] \\ & =\left|[0-(-2)]-\left[\left(\frac{-9}{2}\right)-(-4)\right]+\left[(-4)-\left(-\frac{9}{2}\right)\right]\right| \\ & =\left|2+\frac{9}{2}-4-4+\frac{9}{2}\right|=[2+9-8]=3\end{aligned}$

$\begin{aligned} & \text { (c) Let } I=\int_{-1 / 2}^{1 / 2}\left\{[x]+\log \left(\frac{1+x}{1-x}\right) d x\right\} \\ & \because \int_{-a}^a f(x) d x=\left\{\begin{array}{cr} 2 \int_0^a f(x) d x, & f(-x)=f(x) \\ 0, & f(-x)=-f(x) \end{array}\right. \\ & =\int_{-1 / 2}^{1 / 2}[x] d x+\int_{-1 / 2}^{1 / 2} \log \left(\frac{1+x}{1-x}\right) d x \end{aligned}$

Let $f(x)=\log \left(\frac{1+x}{1-x}\right)$

$\begin{aligned} & f(-x)=\log \left(\frac{1-x}{1+x}\right)=-f(x) \quad \quad \text { [odd function] } \\ & =\int_{-1 / 2}^{1 / 2}[x] d x+0=\int_{-1 / 2}^0[x] d x+\int_0^{1 / 2}[x] d x \\ & =\int_{-1 / 2}^0(-1) d x+\int_0^{1 / 2} 0 d x \\ & =-(x)_{-1 / 2}^0+0=-\left(0+\frac{1}{2}\right)=-\frac{1}{2} \quad \end{aligned}$