Definite Integration

$ \begin{aligned} &\text { We have, } P=\lim _{x \rightarrow \infty}\\ &\begin{aligned} & \quad\left[\left(1+\frac{1}{n^3}\right)^{1 / n^3}-\left(1+\frac{8}{n^3}\right)^{4 / n^3} \ldots(2)^{1 / n}\right] \\ & \Rightarrow \quad \log p=\lim _{x \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{r^2}{n^2} \log \left(1+\frac{r^3}{n^3}\right) \\ & \Rightarrow \quad \log p=\int_0^1 x^2 \log \left(1+x^3\right) d x \\ & \text { Put } 1+x^3=t \Rightarrow 3 x^2 d x=d t \\ & \text { when } n=0, t=1 \text { and } x=1, t=2 \\ & \therefore \log p=\frac{1}{3} \int_1^2 \log t d t=\frac{1}{3}[t \log t-t]_1^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \log p=\frac{1}{3}[2 \log 2-1] \\ & \text { Hence, } p=e^{(2 \log 2-1 / 3)} \end{aligned} $

To solve the integral $ \int_{-5 \pi}^{5 \pi}(1-\cos 2 x)^{\frac{5}{2}} d x $, we start by simplifying the expression inside the integral:

Notice that $ 1 - \cos 2x = 2 \sin^2 x $. Substituting this in, the integral becomes:

$ I = \int_{-5 \pi}^{5 \pi} (1 - \cos 2x)^{\frac{5}{2}} \, dx = \int_{-5 \pi}^{5 \pi} (2 \sin^2 x)^{\frac{5}{2}} \, dx $

Simplify the expression:

$ = \int_{-5 \pi}^{5 \pi} 2^{\frac{5}{2}} \cdot |\sin^5 x| \, dx $

Since $|\sin^5 x| = \sin^5 x$ for the interval $[0, \pi]$ and is symmetrical over $[-\pi, 0]$, the integral becomes:

$ = 8 \sqrt{2} \int_0^{5\pi} \sin^5 x \, dx $

Exploit the periodicity of the sine function to break the integral over one full period:

$ = 40 \sqrt{2} \int_0^{\pi} \sin^5 x \, dx $

Using the reduction formula for $\int_0^\pi \sin^n x \, dx$, the final result of this reduction gives:

$ = \frac{128 \sqrt{2}}{3} $

$I=\int_0^{\pi / 4} \log (1+\tan x) d x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Using $\int_0^a f(x) d x=\int_0^a f(a-x) d x$

$ \begin{aligned} I & =\int_0^{\pi / 4} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\ & =\int_0^{\pi / 4} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x \\ & \left(\because \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\right) \\ I & =\int_0^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

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

$ \begin{aligned} 2 I= & \int_0^{\pi / 4} \log (1+\tan x) d x \\ & \quad+\int_0^{\pi / 4} \log \frac{2}{(1+\tan x)} d x \\ = & \int_0^{\pi / 4} \log (1+\tan x) \cdot \frac{2}{(1+\tan x)} d x \\ = & \int_0^{\pi / 4} \log 2 d x=\log 2 \int_0^{\pi / 4} d x \\ = & \log 2 \cdot(x)_0^{\pi / 4}=\log 2 \cdot\left(\frac{\pi}{4}-0\right) \\ 2 I= & \frac{\pi}{4} \log 2 \Rightarrow I=\frac{\pi}{8} \log 2 \end{aligned} $

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

$ \begin{aligned} &\begin{aligned} & =2\left[\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x\right] \\ & =2\left[\pi \int_0^\pi \frac{\sin x d x}{1+\cos ^2 x}-\int_0^\pi \frac{x \sin x}{1+\cos ^2 x}\right] \\ 2 I & =2 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x \end{aligned}\\ &\text { }\\ &\begin{aligned}Let\,\,\,\,\, t & =\cos x \\ d t & =-\sin x d x \\ & =-\pi \int_1^{-1} \frac{1}{1+t^2} d t=\pi \int_{-1}^1 \frac{1}{1+t^2} d t \\ I & =\pi\left[\tan ^{-1}(t)\right]_{-1}^1 \\ & =\pi\left[\frac{\pi}{4}+\frac{\pi}{4}\right]=\frac{\pi^2}{2} \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} & \quad I=\int_0^{\pi / 4} \frac{x^2 d x}{(x \sin x+\cos x)} \Rightarrow I=\int_0^{\pi / 4} \frac{x \sec x \cdot \cos x \cdot d x}{(x \sin x+\cos x)^2} \\ & \Rightarrow \quad I=\int_0^{\pi / 4} x \sec x \cdot \frac{x \cos x}{(x \sin x+\cos x)^2} d x \end{aligned} $

Let $I_1=x \sec x$ and $I_2=\frac{x \cos x}{x \sin x+\cos x}$

Let $x \sin x+\cos x=t$

$ \begin{aligned} & x \cos x+\sin x-\sin x=\frac{d t}{d x} \\ & x \cos x d x=d t \end{aligned} $

$ \begin{array}{ll} \Rightarrow & I_2=\int \frac{x \cos x}{(x \sin x+\cos x)^2}=\int \frac{d t}{t^2}=-\frac{1}{t} \\ \Rightarrow \quad & I_2=\frac{-1}{(x \sin x+\cos x)} \\ \Rightarrow \quad & I=\int_0^{\pi / 4} x \sec x \cdot \frac{x \cos x}{(x \sin x+\cos x)} d x \\ \Rightarrow \quad & I=\left[x \sec x \cdot \frac{-1}{(x \sin x+\cos x)}\right]_0^{\frac{\pi}{4}}- \int_0^{\pi / 4}[\sec x+x \sec x \tan x] \cdot \frac{-1}{(x \sin x+\cos x)} d x \\ \because \quad & \int f(x) g(x) d x=f(x) \int g(x) d x-\int\left[\frac{d}{d x} f(x)-\int g(x) d x\right] d x \end{array} $

$ \begin{array}{ll} \Rightarrow & I=\left[\frac{-x \sec x}{x \sin x+\cos x}\right]_0^{\pi / 4}+\int_0^{\pi / 4} \frac{\sec x(1+x \tan x) d x}{(x \sin x+\cos x)} \\ \Rightarrow & I=\left[\frac{-x \sec x}{x \sin x+\cos x}\right]_0^{\pi / 4}+\int_0^{\pi / 4} \frac{\sec ^2 x(\cos x+x \sin x)}{(x \sin x+\cos x)} d x \\ \Rightarrow & I=\left[\frac{-x \sec x}{(x \sin x+\cos x)}\right]_0^{\pi / 4}+\int_0^{\pi / 4} \sec ^2 x d x \end{array} $

$ \begin{array}{ll} \Rightarrow & I=\frac{-\frac{\pi}{4} \cdot \sqrt{2}}{\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}+\int_0^{\pi / 4} \sec ^2 x d x \\ \Rightarrow & I=\frac{-\pi \sqrt{2} / 4}{(\pi+4) / 4 \sqrt{2}}+[\tan x]_0^{\pi / 4} \\ \Rightarrow & I=\frac{-2 \pi}{\pi+4}+(1-0)=\frac{-2 \pi}{\pi+4}+1 \Rightarrow I=\frac{4-\pi}{\pi+4} \end{array} $

We have, $I=\int_0^1 \frac{x}{(1-x)^{\frac{3}{4}}} d x$

Let $1-x=t \Rightarrow x=1-t \Rightarrow-d x=d t$

$ \begin{aligned} & \therefore \quad I=\int_0^1 \frac{1-t}{t^{3 / 4}} d t \Rightarrow I=\int_0^1 t^{-3 / 4}(1-t) d t \\ & \Rightarrow \quad I=\int_0^1 t^{-3 / 4} d t-\int_0^1 t^{1 / 4} d t \Rightarrow I=\left[4 t^{1 / 4}\right]_0^1-\left[\frac{4 t^{5 / 4}}{5}\right]_0^1 \end{aligned} $

Putting the value of $t$

$ I=[4(1-x)]_0^1+\left[\frac{4}{5}(1-x)^{5 / 4}\right]_0^1 $

On solving, we get

$ I=\frac{16}{5} $

We have,

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

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

$ \begin{aligned} & \Rightarrow f(-x)=\sqrt{1-x+x^2}-\sqrt{1+x+x^2} \\ & \Rightarrow f(-x)=-\left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) \Rightarrow f(-x)=-f(x) \end{aligned} $

$\because f(x)$ is an odd function.

As, we know that

$ \int_{-a}^a f(x) d x=\left\{\begin{array}{l} \int_0^{2 a} f(x) d x, \text { if } f(-x)=f(x) \text {, even function } \\ 0, \text { if } f(-x)=-f(x), \text { odd funciton } \end{array}\right\} $

$ \begin{aligned} & \text { Then, } I=\int_{-1}^1 \sqrt{1+x+x^2} \\ & \qquad-\sqrt{1-x+x^2} d x=0 \end{aligned} $

We have,

$ \Rightarrow \quad I=\int_1^5|x-3|+|1-x| d x $

Let $f(x)=|x-3|+|1-x|$

Here, when $1 \leq x \leq 3$

$ \Rightarrow f(x)=-x-3-1+x=2 $

When, $3 \leq x \leq 5$

$ \begin{aligned} \Rightarrow f(x) & =x-3-1+x=2 x-4 \\ \Rightarrow \quad I & =\int_1^5|x-3|+|1-x| d x \\ & =\int_1^3 2 d x+\int_3^5(2 x-4) d x \\ \Rightarrow \quad I & =[2 x]_1^3+\left[\frac{2 x^2}{2}-4 x\right]_3^5 \end{aligned} $

On solving, we get

$ I=12 $

If $729 \int_1^3 \frac{1}{x^3\left(x^2+9\right)^2} d x=a+\log b$  Then, $a-b$

On substituting $u=x^2$

$ \begin{array}{ll} \Rightarrow & d u=2 x d x \\ \Rightarrow & d x=\frac{1}{2} \frac{d u}{\sqrt{u}} \\ \text { when, } x=1 & \end{array} $

$ \begin{array}{ll} \Rightarrow & u=1 \\ & x=3 \\ \Rightarrow & u=9 \end{array} $

$ \begin{aligned} &\begin{aligned} I & =729 \int_1^9 \frac{1}{u^{\frac{3}{2}}(u+9)^2} d u \\ & =729 \int_1^9 \frac{1}{2 u^{\frac{3}{2}+\frac{1}{2}}(u+9)^2} d u \\ & =729 \int_1^9 \frac{1}{2 u^2(u+9)^2} d u \end{aligned}\\ &\text { Partial fraction }\\ &\frac{1}{u^2(u+9)^2}=\frac{A}{u}+\frac{B}{u^2}+\frac{C}{(u+9)}+\frac{D}{(u+9)} \end{aligned} $

$ \begin{aligned} &\begin{gathered} \frac{1}{u^2(u+9)^2} \frac{A\left(u(u+9)^2+B(u+9)^2+C u^2(u+9)+D u^2\right.}{u^2(u+9)^2} \\ 1=A u\left(u^2+81+18 u\right)+B\left(u^2+81+18 u\right)+C u^3+9 C u^2+D u^2 \\ 1=A u^3+81 A u+18 A u^2+B u^2+81 B+18 u B+C u^3+9 C u^2+D u^2 \\ 1=(A+C) u^3+(18 A+B+9 C+D) u^2+(81 A+18 B) u+81 B \end{gathered}\\ &\text { Comparison of coefficient } u^3, u^2, u \text { and } 1 \text {. }\\ &\begin{aligned} & A+C=0,18 A+B+9 C+D=0 \\ & 81 A+18 B=0 \Rightarrow 81 B=1 \\ & B=\frac{1}{81} \Rightarrow 81 A+18 \cdot \frac{1}{81}=0 \\ & 81 A=-\frac{2}{9} \Rightarrow A=\frac{-2}{729} \\ & \frac{-2}{729}+C=0 \Rightarrow C=\frac{2}{729} \\ & 18\left(\frac{-2}{729}\right)+\frac{1}{81}+9\left(\frac{2}{729}\right)+D=0 \\ & -\frac{4}{81}+\frac{1}{81}+\frac{2}{81}+D=0 \Rightarrow \frac{-1}{81}+D=0 \end{aligned} \end{aligned} $

$ \begin{aligned} D & =\frac{1}{81} \\ & =\frac{729}{2} \int_1^9\left(\left(\frac{-2}{729}\right) \frac{1}{u}+\frac{1}{81 u^2}+\frac{2}{729(u+9)}+\frac{1}{81(u+9)^2}\right) d u \\ & =\frac{729}{2}\left[\frac{-2}{729} \log u+\frac{1}{81} \frac{u^{-2+1}}{(-2+1)}+\frac{2}{729} \log (u+9)+\frac{1}{81} \frac{(u+9)^{-2+1}}{-2+1}\right]_1^9 \\ & =\frac{729}{2}\left[\frac{-2}{729}(\log 9-\log 1)+\frac{1}{81}(-1)\left(\frac{1}{9}-1\right)+\frac{2}{729}[\log (18)-\log (10)]\right. \\ & =\frac{729}{2}\left[\frac{-2}{729} \log 9-\frac{1}{81}\left(\frac{-8}{9}\right)+\frac{2}{729} \log \left(\frac{18}{10}\right)-\frac{1}{81}\left(\frac{-8}{180}\right)\right]+C \\ & =\frac{1}{2}\left[-2 \log 9+8+2 \log \left(\frac{1}{5}\right)+\frac{8}{20}\right]+C \\ & =\frac{1}{2}\left[-2 \log 9+8+2 \log 9-2 \log 5+\frac{2}{5}\right]+C \\ & =\frac{1}{2}\left[8+\frac{2}{5}-2 \log 5\right]+C \\ & =\frac{1}{2}\left[\frac{40+2}{5}-2 \log 5\right] \Rightarrow \frac{42}{2 \times 5}-\log 5 \end{aligned} $

$ \begin{aligned} a & =\frac{42}{2 \times 5}, b=\frac{1}{5} \\ a & =\frac{21}{5}, b=\frac{1}{5} \\ a-b & =\frac{21}{5}-\frac{1}{5}=\frac{20}{5}=4 \end{aligned} $

To find the limit

$ \lim\limits_{n \rightarrow \infty} \frac{1^{77}+2^{77}+\ldots+n^{77}}{n^{78}} $

we use the method of integral approximation to approximate the sum of powers. Consider the sum

$ S_n = 1^{77} + 2^{77} + \ldots + n^{77}. $

This can be approximated using the integral

$ S_n \approx \int_1^n x^{77} \, dx. $

Evaluating this integral, we have

$ \int_1^n x^{77} \, dx = \left[\frac{x^{78}}{78}\right]_1^n = \frac{n^{78}}{78} - \frac{1}{78}. $

As $ n $ becomes very large, the term $\frac{1}{78}$ becomes negligible, and we approximate

$ S_n \approx \frac{n^{78}}{78}. $

Thus, the given limit can be rewritten and evaluated as follows:

$ \lim_{n \rightarrow \infty} \frac{S_n}{n^{78}} = \lim_{n \rightarrow \infty} \frac{\frac{n^{78}}{78}}{n^{78}} = \lim_{n \rightarrow \infty} \frac{1}{78} = \frac{1}{78}. $

Therefore, the value of the limit is

$ \frac{1}{78}. $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } f(x)=\left\{\begin{array}{cc} \frac{6 x^2+1}{4 x^3+2 x+3}, & 0< x < 1 \\ x^2+1 & 1 \leq x < 2 \end{array}\right. \\ & \int_0^2 f(x) d x=\int_0^1\left(\frac{6 x^2+1}{4 x^3+2 x+3}\right) d x+\int_1^2\left(x^2+1\right) d x \\ & \text { Let } u=4 x^3+2 x+3 \\ & \qquad d u=\left(12 x^2+2\right) d x \Rightarrow 2\left(6 x^2+1\right) d x \\ & \Rightarrow d x=\frac{d u}{2\left(6 x^2+1\right)} \\ & \quad x=0 \Rightarrow u=3 \\ & \quad x=1 \Rightarrow u=4+2+3=9 \\ & \text { Or }\left(6 x^2+1\right) d x=\frac{d u}{2} \Rightarrow \int_3^9 \frac{1}{u}\left(\frac{d u}{2}\right)+\int_1^2\left(x^2+1\right) d x \end{aligned} \end{aligned} $

$ \begin{aligned} & \frac{1}{2}[\log u]_3^9+\left[\frac{x^3}{3}+x\right]_1^2 \\ & \frac{1}{2}[\log 9-\log 3]+\left[\frac{8}{3}+2-\frac{1}{3}-1\right] \\ & \frac{1}{2}\left[\log \left(\frac{9}{3}\right)+\left(\frac{8+6}{3}-\frac{4}{3}\right)\right] \\ & \frac{1}{2} \log 3+\frac{14}{3}-\frac{4}{3}=\frac{1}{2} \log 3+\frac{10}{3} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \int_1^n[x] d x=120 \\ & =\int_1^2 1^{d x}+\int_2^3 2^{d x}+\ldots+\int_{n-1}^n(n-1) d x=120 \\ & \Rightarrow \quad[x]_1^2+[2 x]_2^3+\ldots[(n-1) x]_{n-1}^n=120 \\ & \Rightarrow \quad 2-1+6-4+\ldots+(n-1)=120 \\ & \Rightarrow \quad 1+2+\ldots+n-1=120 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{n(n-1)}{2}=120 \\ & \Rightarrow \quad n^2-n-240=0 \\ & \Rightarrow \quad n^2-16 n+15 n-240=0 \\ & \Rightarrow \quad n(n-16)+15(n-16)=0 \\ & (n-16)(n+15)=0 \\ & n=16 \end{aligned} $

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

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} $

$ \begin{aligned} &\text { We have } I=\int\limits_{-3}^3|2-x| d x\\ &\begin{aligned} & |2-x|= \begin{cases}(2-x) & -3 < x < 2 \\ (x-2) & 2 \leq x<3\end{cases} \\ & \int\limits_{-3}^3|2-x| d x=\int\limits_{-3}^2(2-x) d x+\int\limits_2^3(x-2) d x \\ & =\left[2 x-\frac{x^2}{2}\right]_{-3}^2+\left[\frac{x^2}{2}-2 x\right]_2^3 \\ & =(4-2)-\left(-6-\frac{9}{2}\right)+\left(\frac{9}{2}-6\right)-(2-4) \\ & =2+6+\frac{9}{2}+\frac{9}{2}-6+2=13 \end{aligned} \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} $

1. The given integral is:

$ I=\int_0^1 \frac{dx}{\left(5+2x-2x^2\right)\left(1+e^{2-4x}\right)} $ .............(i)

1. Perform a substitution, $x \rightarrow 1-x$. This gives:

$ I=\int_0^1 \frac{dx}{\left(5+2(1-x)-2(1-x)^2\right)\left(1+e^{2-4(1-x)}\right)} $

1. Simplify the expression inside the integral:

$ I=\int_0^1 \frac{dx}{\left(5+2-2x-2(1-2x+x^2)\right)\left(1+e^{4x -2}\right)} $ ............(ii)

1. Add the original integral (i) and the integral after substitution (ii):

$ 2I=\int_0^1 \frac{dx}{5+2x-2x^2} $

1. Factor the quadratic expression in the denominator:

$ 2I=\int_0^1 \frac{dx}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)} $

1. To solve this integral, we can perform a change of variables using the substitution
   
   $x-\frac{1}{2}= \frac{\sqrt{11}}{2}\tan{u}$, then $dx=\frac{\sqrt{11}}{2}\sec^2{u} du$:

$ I=\frac{1}{\sqrt{11}}\int \frac{\sec^2{u}}{1+\tan^2{u}}du $

1. Using the identity $\sec^2{u}=1+\tan^2{u}$:

$ I=\frac{1}{\sqrt{11}}\int du = \frac{1}{\sqrt{11}}(u+C) $

1. Now we need to find the limits of the integral after the substitution. If $x=0$, then $u=\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)$. If $x=1$, then $u=\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)$. So, the integral becomes:

$ I=\frac{1}{\sqrt{11}}\left[\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)-\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)\right] $

1. Using the properties of the arctangent function, we can rewrite the integral as:

$ I=\frac{1}{\sqrt{11}} \ln\left(\frac{\sqrt{11}+1}{\sqrt{10}}\right) $

1. From this result, we have $\alpha=\sqrt{11}$ and $\beta=\sqrt{10}$. Now, we can find $\alpha^4 - \beta^4$:

$ \alpha^4 - \beta^4 = (11)^2 - (10)^2 = 121 - 100 = 21 $

Thus, $\alpha^4 - \beta^4 = 21$.

We're given the expression:

$\frac{e^{-\pi/4} + \int_0^{\pi / 4} e^{-x} \tan ^{50} x dx}{\int_0^{\pi / 4} e^{-x}(\tan x)^{49} dx + \int_0^{\pi / 4} e^{-x}(\tan x)^{51} dx}$

Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x} \tan^n x$ from 0 to $\pi / 4$, where $n$ is an integer. Let's denote this integral as $I(n)$:

$I(n) = \int_0^{\pi / 4} e^{-x} \tan^n x dx$

We can then rewrite the original expression in terms of $I(n)$:

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)}$

Now, we'll apply the method of integration by parts, which states that for two functions $u(x)$ and $v(x)$:

$\int u dv = uv - \int v du$

We'll choose:

$u = \tan^n x, \quad dv = e^{-x} dx$

Then we get:

$du = n \tan^{n-1} x \sec^2 x dx, \quad v = -e^{-x}$

Applying integration by parts, we have:

$I(n) = -e^{-x} \tan^n x \Bigg|_0^{\pi / 4} + n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x \sec^2 x dx$

Since $\tan(\pi / 4) = 1$, the first term evaluates to:

$-e^{-\pi / 4}$

The second term becomes:

$n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x (1 + \tan^2 x) dx = n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x dx + n \int_0^{\pi / 4} e^{-x} \tan^{n+1} x dx$

This is equal to:

$n(I(n-1) + I(n+1))$

So we have:

$I(n) = -e^{-\pi / 4} + n(I(n-1) + I(n+1))$

Now we can substitute $n = 50$ into this equation:

$I(50) = -e^{-\pi / 4} + 50(I(49) + I(51))$

So the original expression becomes:

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)} = \frac{e^{-\pi/4} - e^{-\pi / 4} + 50(I(49) + I(51))}{I(49) + I(51)} = 50$

$ \begin{aligned} & S_1: \lim _{n \rightarrow \infty} \frac{1}{n^2}[2+4+6+\ldots+2 n] \\\\ & \lim _{n \rightarrow \infty} 2 \frac{n(n+1)}{2 n^2}=1 \\\\ & S_2: \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(\sum r^{15}\right)=\lim _{n \rightarrow \infty} \frac{1}{n} \sum\left(\frac{r}{n}\right)^{15} \\\\ & =\int_0^1 x^{15} d x=\frac{1}{16} \\\\ & \therefore \text { Both } S_1 \text { and } S_2 \text { are correct. } \end{aligned} $

$ \begin{aligned} & \mathrm{l}=\int_0^{\infty} \frac{6}{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+2\right)\left(\mathrm{e}^{\mathrm{x}}+3\right)} \mathrm{dx} \\\\ & =6 \int_0^{\infty}\left(\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+1}+\frac{-1}{\mathrm{e}^{\mathrm{x}}+2}+\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+3}\right) \mathrm{dx} \\\\ & =3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+\mathrm{e}^{-\mathrm{x}}} \mathrm{dx}-6 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}} \mathrm{dx}}{1+2 \mathrm{e}^{-\mathrm{x}}}+3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+3 \mathrm{e}^{-\mathrm{x}}} \mathrm{dx} \\\\ & =3\left[-\ln \left(1+\mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty}+6 \frac{1}{2}\left[\ln \left(1+2 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} -\frac{3}{3}\left[\ln \left(1+3 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} \\\\ & =3 \ln 2-3 \ln 3+\ln 4 \\\\ & =3 \ln \frac{2}{3}+\ln 4 \\\\ & =\ln \frac{32}{27} \end{aligned} $

The integral equation is given by :

$\int\limits_0^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$

Step 1 : Break the first integral into two parts :

$I = \int\limits_0^{\frac{\pi}{4}} f(\sin 2x) \sin x \, dx + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx$

3. Apply the King's property, $\int\limits_a^b f(x) dx = \int\limits_a^b f(a+b-x) dx$, to the first integral and substitute $x - \frac{\pi}{4} = t$ in the second integral.

This gives :

$ \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin(\frac{\pi}{4}-x) \, dx + \int\limits_0^{\frac{\pi}{4}} f(\cos 2t) \sin(\frac{\pi}{4}+t) \, dt + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0 $

$ \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) [2 \sin \frac{\pi}{4} \cos x + \alpha \cos x] \, dx = 0 $

Then, noticing that $2 \sin \frac{\pi}{4} = \sqrt{2}$, you can factor out the term $\cos x$:

$ = (\alpha + \sqrt{2}) \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0 $

In order for this equation to hold true, either the integral of the function is zero, or the term outside the integral is zero. Since we have no reason to assume that the integral of the function is zero, we set the term outside the integral to zero, yielding the solution:

$ \alpha + \sqrt{2} = 0 \Rightarrow \alpha = -\sqrt{2} $

So, the correct answer to the original problem is $\alpha = -\sqrt{2}$, which corresponds to Option C.

$ \begin{aligned} \operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}\right\} & =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\\\ {\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}\right] } & =1 ; \mathrm{x} \in[1,2) \end{aligned} $

$ \therefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \\\\ \mathrm{e} ; \mathrm{x} \in[1,2) \end{array}\right. $

$ \begin{aligned} & \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\\\ & x^2=t \Rightarrow 2 x d x=d t \end{aligned} $

$ =\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}\right|_1 ^2 $

$ \begin{aligned} & =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\\\ & =2 \mathrm{e}-\frac{1}{2} \end{aligned} $

$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$

Let $e^x=t \Rightarrow e^x d x=d t$

When, $x \rightarrow-\log _e 2$, then $t \rightarrow \frac{1}{2}$

When, $x \rightarrow \log _e 2$, then $t \rightarrow 2$

$ I=\int_\limits{\frac{1}{2}}^2\left[\log _e\left(t+\sqrt{1+t^2}\right)\right] d t $ ...........(i)

On applying integration by part method in Eq. (i), we get

$ \begin{aligned} & I=\left[t \log _e\left(t+\sqrt{1+t^2}\right)\right]_{1 / 2}^2-\int_{1 / 2}^2 \frac{t}{t+\sqrt{1+t^2}}\left(1+\frac{2 t}{2 \sqrt{1+t^2}}\right) d t \\\\ & =2 \log _e(2+\sqrt{5})-\frac{1}{2} \log _e\left(\frac{1+\sqrt{5}}{2}\right)-\int_{1 / 2}^2 \frac{t}{\sqrt{1+t^2}} d t \end{aligned} $

$ =\log _e\left(\frac{(2+\sqrt{5})^2}{\left(\frac{1+\sqrt{5}}{2}\right)^{1 / 2}}\right)-\frac{1}{2} \int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t $ .............(ii)

Let $\quad I_1=\int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t$

Let $1+t^2=w$

$2 t d t=d w$

When, $t \rightarrow \frac{1}{2}$, then $w=\frac{5}{4}$

When, $t \rightarrow 2$, then $w=5$

$ \begin{aligned} I_1 & =\int_{5 / 4}^5 \frac{1}{\sqrt{w}} d w \\\\ & =[2 \sqrt{w}]_{5 / 4}^5 \\\\ & =2\left[\sqrt{5}-\frac{\sqrt{5}}{2}\right]=\sqrt{5} \end{aligned} $

On substitute value of $I_1$ in Eq. (ii), we get

$ I=\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2} $

Given that

$ \int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0 $

On differentiating using Leibnitz rule, we get

$ \begin{aligned} & \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\ & \Rightarrow f\left(t^2\right)+t^4=2 t \\\\ & \Rightarrow f\left(t^2\right)=2 t-t^4 \end{aligned} $

On substituting $\frac{\pi}{2}$ for $t$, we get

$ f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right) $

Let $I=\int\limits_0^\pi f(x) \sin x d x$ ..........(i)

$ \begin{aligned} & =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\\\ & =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii) \end{aligned} $

On adding Equations (i) and (ii), we get

$ \begin{aligned} & 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\\\ & \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\\\ & =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\\\ & \Rightarrow I=\pi^2 \end{aligned} $

Let $L=\lim\limits_{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)\right\}$

Here, $\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{3}}-2^{\frac{1}{5}}\right)$ $ \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)<\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $

$ \Rightarrow \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $

Since, $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0$ and $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n=0$

$\therefore L = 0$

We have,

$ 5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x>0 $ ..........(i)

On replacing $x$ by $\frac{1}{x}$ in (i), we get

$ 5 f\left(\frac{1}{x}\right)+4 f(x)=x+3 $ ..........(ii)

Now, using Eq. (i) $\times 5-$ (ii) $\times 4$, we get

$ \begin{aligned} & 25 f(x)-16 f(x) =\left(\frac{5}{x}+15\right)-(4 x+12) \\\\ &\Rightarrow 9 f(x) =\frac{5}{x}-4 x+3 \\\\ &\Rightarrow f(x) =\frac{1}{9}\left(\frac{5}{x}-4 x+3\right) ...........(iii) \end{aligned} $

$ \begin{aligned} \therefore \quad & 18 \int_1^2 f(x) d x=18 \int_1^2 \frac{1}{9}\left(\frac{5}{x}-4 x+3\right) d x \text { [Using Eq. (iii)] }\\\\ = & 2 \int_1^2\left(\frac{5}{x}-4 x+3\right) d x \end{aligned} $

$ \begin{aligned} & =2\left[5 \log _e x-4\left(\frac{x^2}{2}\right)+3 x\right]_1^2 \\\\ & =2\left[\left(5 \log _e 2-2(2)^2+3(2)\right)-\left(5 \log 1-2(1)^2+3(1)\right)\right] \\\\ & =2\left[5 \log _e 2-8+6+2-3\right] [\because \log 1=0]\\\\ & =10 \log _e 2-6 \end{aligned} $

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

Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

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

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

Now,

$ \begin{aligned} & \tan x=t \\\\ & =\frac{\pi}{2} \int_0^1 \frac{d t}{1+3 t^2} \\\\ & =\frac{\pi}{2}\left[\frac{\tan ^{-1}(\sqrt{3} t)}{\sqrt{3}}\right]_0^1 \\\\ & =\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}} \end{aligned} $

$ \begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots \ldots+\frac{1}{2 n}\right] \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{r+n} \\\\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{1}{\frac{r}{n}+1}\right) \\\\ & =\int_0^1 \frac{d x}{x+1} \\\\ & =\left.\log _e(1+\mathrm{x})\right|_0 ^1 \\\\ & =\log _e^2 \end{aligned} $

$I=\int\limits_{0}^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x, \alpha>0$

$ \begin{aligned} & =\frac{1}{\alpha} \int\limits_{0}^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) d x \end{aligned} $

$\eqalign{ & = {1 \over \alpha }\int\limits_0^\alpha {\left( {x\sqrt {x + \alpha } + x\sqrt x } \right)} dx \cr & = {1 \over \alpha }\int\limits_0^\alpha {\left[ {\left( {x + \alpha - \alpha } \right)\sqrt {x + \alpha } + {x^{{3 \over 2}}}} \right]} dx \cr} $

$ \begin{aligned} & =\frac{1}{\alpha}\int\limits_0^\alpha \left[(x+\alpha)^{3 / 2}-\alpha(x+\alpha)^{1 / 2}+x^{3 / 2}\right] d x \\\\ & =\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5 / 2}-\alpha \frac{2}{3}(x+\alpha)^{3 / 2}+\frac{2}{5} x^{5 / 2}\right]_0^\alpha \\\\ & =\frac{1}{\alpha}\left(\frac{2}{5}(2 \alpha)^{5 / 2}-\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+\frac{2}{5} \alpha^{5 / 2}-\frac{2}{5} \alpha^{5 / 2}+\frac{2}{3} \alpha^{5 / 2}\right) \\\\ & =\frac{1}{\alpha}\left(\frac{2^{7 / 2} \alpha^{5 / 2}}{5}-\frac{2^{5 / 2} \alpha^{5 / 2}}{3}+\frac{2}{3} \alpha^{5 / 2}\right)=\alpha^{3 / 2}\left(\frac{2^{7 / 2}}{5}-\frac{2^{5 / 2}}{3}+\frac{2}{3}\right) \\\\ & =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10)=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10) \end{aligned} $

Now, $\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)=\frac{16+20 \sqrt{2}}{15}=\frac{2 \sqrt{2}(4 \sqrt{2}+10)}{15}$

$ \Rightarrow \alpha^{3 / 2}=2 \sqrt{2}=2^{3 / 2} $

$\Rightarrow \alpha=2 $

$\phi(x)=\frac{1}{\sqrt{x}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$

$\Rightarrow \phi^{\prime}(x)=\frac{-1}{2 x^{3 / 2}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$

$+\frac{1}{\sqrt{x}}\left(4 \sqrt{2} \sin (x)-3 \phi^{\prime}(x)\right)$

At, $x=\frac{\pi}{4}$

$\phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{-1}{2\left(\frac{\pi}{4}\right)^{3 / 2}} \times 0+\sqrt{\frac{4}{\pi}}\left(4 \sqrt{2} \times \frac{1}{\sqrt{2}}-3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)$

$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)\left[1+\frac{6}{\sqrt{\pi}}\right]=\frac{2}{\sqrt{\pi}} \times 4$

$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{8}{6+\sqrt{x}}$

$\int_{0}^{\alpha} \frac{t^{50}}{1-t} d t$

$ \begin{aligned} & = \int_{0}^{\alpha} \frac{t^{50}-1+1}{1-t}\\\\ & =-\int_{0}^{\alpha}\left(\frac{1-t^{50}}{1-t}-\frac{1}{1-t}\right) d t\\\\ & =-\int_{0}^{\alpha}\left(1+t+\ldots . .+t^{49}\right)+\int_{0}^{\alpha} \frac{1}{1-t} d t \end{aligned} $

$=-\left(\frac{\alpha^{50}}{50}+\frac{\alpha^{49}}{49}+\ldots . .+\frac{\alpha^{1}}{1}\right)+\left(\frac{\ln (1-\mathrm{f})}{-1}\right)_{0}^{\alpha}$

$=-\mathrm{P}_{50}(\alpha)-\ln (1-\alpha)$

$=-\mathrm{P}_{50}(\alpha)-\beta$

Let I = $\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$

$ =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin x(1+\cos x)} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x $

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

$ =\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin ^3 x} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2 \cos ^2 \frac{x}{2}} d x $

$ \begin{array}{r} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin ^3 x} d x-\int\limits_{\pi / 3}^{\pi / 2} 2 \cot x \operatorname{cosec}^2 x d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2} \sec ^2 \frac{x}{2} d x \end{array} $

$ =2 I_1-2 I_2+\frac{3}{2} I_3 $

Now,

$ \begin{aligned} I_1 & =\int\limits_{\pi / 3}^{\pi / 2} \operatorname{cosec}^3 x d x \\\\ & =\int\limits_{\pi / 3}^{\pi / 2} \sqrt{1+\cot ^2 x} \operatorname{cosec}^2 x d x \end{aligned} $

Put $\cot x=t \Rightarrow-\operatorname{cosec}^2 x d x=d t$

When, $x=\frac{\pi}{3} \Rightarrow t=\frac{1}{\sqrt{3}} \text { and } x=\frac{\pi}{2} \Rightarrow t=0$

$ \begin{aligned} \therefore & I_1=-\int\limits_{\frac{1}{\sqrt{3}}}^0 \sqrt{1+t^2} d t \\\\ = & \int\limits_0^{\frac{1}{\sqrt{3}}} \sqrt{1+t^2} d t \\\\ = & {\left[\frac{t}{2} \sqrt{1+t^2}+\frac{1}{2} \log \left[t+\sqrt{1+t^2}\right]\right]_0^{\frac{1}{\sqrt{3}}} } \\\\ = & \frac{1}{3}+\frac{1}{2} \log \sqrt{3} \end{aligned} $

$ I_2=\int\limits_{\pi / 3}^{\pi / 2} \cot x \operatorname{cosec}^2 x d x $

Put $\cot x=t \Rightarrow \operatorname{cosec}^2 x d x=-d t$

When, $x=\frac{\pi}{3}$ $\Rightarrow t=\frac{1}{\sqrt{3}}$ and $x=\frac{\pi}{2} \Rightarrow t=0$

$ \therefore \begin{aligned} I_2 & =-\int\limits_{\frac{1}{\sqrt{3}}}^0 t d t \\\\ & =\int\limits_0^{\frac{1}{\sqrt{3}} t} t d t=\left[\frac{t^2}{2}\right]_0^{\frac{1}{\sqrt{3}}}=\frac{1}{6} \end{aligned} $

$ \begin{aligned} & I_3=\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec ^2 \frac{x}{2} d x \\\\ = & 2\left(\tan \frac{\pi}{4}-\tan \frac{\pi}{6}\right)=2\left(1-\frac{1}{\sqrt{3}}\right) \end{aligned} $

$ \begin{aligned} & \therefore \quad I=2 I_1-2 I_2+\frac{3}{2} I_3 \\\\ & =2\left(\frac{1}{3}+\frac{1}{2} \log \sqrt{3}\right)-2\left(\frac{1}{6}\right) +\frac{3}{2}\left(2\left(1-\frac{1}{\sqrt{3}}\right)\right) \end{aligned} $

$ \begin{aligned} & =\frac{2}{3}+\log \sqrt{3}-\frac{1}{3}+3-\sqrt{3} \\\\ & =\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3} \end{aligned} $

$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{3 \over n}{{\left( {2 + {r \over n}} \right)}^2}} $

$ = \int_0^1 {3{{(2 + x)}^2}\,dx} $

$ = \left. {3\,.\,{{{{(2 + x)}^3}} \over 3}} \right|_0^1$

$ = {3^3} - {2^3} = 19$

$I = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $

$ = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{1 + [{x^3}]}}dx} $ ($\because$ $[x] = 1$ when $x \in (12)$)

$ = 3(e - 1)\int_1^2 {{x^2}{e^{[{x^3}]}}dx} $

Let ${x^3} = t$

$I = (e - 1)\int_1^8 {{e^{[t]}}dt} $

$ = ({e^{ - 1}})({e^1} + {e^2} + {e^3}\, + \,...\, + \,{e^7})$

$ = (e - 1)e{{({e^7} - 1)} \over {e - 1}}$

$ = {e^8} - e$

$\int_1^2 {{{{t^4} + 1} \over {{t^6} + 1}}dt} $

$ = \int_1^2 {{{{{({t^2} + 1)}^2}} \over {{t^6} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{t^6} + 1}}dt} } $

$ = \int_1^2 {{{{t^2} + 1} \over {{t^4} - {t^2} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{{({t^3})}^2} + 1}}dt} } $

$ = \left. {{{\tan }^{ - 1}}(2t + \sqrt 3 ) + {{\tan }^{ - 1}}(2t - \sqrt 3 )} \right|_1^2 - \left. {{2 \over 3}{{\tan }^{ - 1}}({t^3})} \right|_1^2$

$ = {\tan ^{ - 1}}(4 + \sqrt 3 ) + {\tan ^{ - 1}}(4 - \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2\sqrt 3 ) - {2 \over 3}({\tan ^{ - 1}}8 - {\tan ^{ - 1}}1)$

$ = {\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$

$I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $ ..... (i)

$x \to {1 \over x}$

$I = \int\limits_{{1 \over 2}}^2 {{1 \over x}{{\tan }^{ - 1}}{1 \over x}dx} $ ..... (ii)

$2I = \int\limits_{{1 \over 2}}^2 {{1 \over x}\,.\,{\pi \over 2}dx} $

$ = \left. {{\pi \over 2}\ln x} \right|_{{1 \over 2}}^2 = \pi \ln 2$

$ \Rightarrow I = {\pi \over 2}\ln 2$

$f(x)=x+\int\limits_{0}^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$

$f(x)=x+\int\limits_{0}^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x)\quad....(1)$

On comparing with

$f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x, x \in \mathbb{R}$ then

$\Rightarrow \frac{a}{\pi^{2}-4}=\int_{0}^{\pi / 2} \cos y f(y) d y \quad....(2)$

$\Rightarrow \frac{b}{\pi^{2}-4}=\int_{0}^{\pi / 2} \sin y f(y) d y \quad....(3)$

Add (2) and (3)

$\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f(y) d y \quad....(4)$

$\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f\left(\frac{\pi}{2}-y\right) d y\quad....(5)$

Add (4) and (5)

$\frac{2(a+b)}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y)\left(\frac{\pi}{2}+\frac{(a+b)}{\pi^{2}-4}(\sin y+\cos y)\right) d y$

$=\pi+\frac{a+b}{\pi^{2}-4}\left(\frac{\pi}{2}+1\right)$

$(a+b)=-2 \pi(\pi+2)$

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

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

$=\frac{1}{8} \ln \left(x^{2}+2\right)-\frac{\ln x}{4}-\frac{1}{8\left(x^{2}+2\right)}-\frac{1}{8 x^{3}}$

Now, $16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}=2 \ln 6-2 \ln 3-4 \ln 2+\frac{11}{6}$

$=\frac{11}{6}-\ln 4$

$ f(x)=\int_0^2 e^{|x-t|} d t $

For $x>2$

$ f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right) $

For $x<0$

$ f(x)=\int_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right) $

For $x \in[0,2]$

$ \begin{aligned} & f(x)=\int_0^x e^{x-t} d t + \int_x^2 e^{t-x} d t \\\\ & =e^{2-x}+e^x-2 \end{aligned} $

For $x>2$

$ \left.f(x)\right|_{\min =e^2-1} $

For $\mathrm{x}<0$

$ \left.f(x)\right|_{\min =e^2-1} $

For $x \in[0,2]$

$ \left.f(x)\right|_{\min }=2(e-1) $

$ \int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x $

We have $\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C$

Hence $\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x=\frac{48}{2} \times\left[\sin ^{-1} \frac{2 x}{3}\right]_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}}$

$=24 \times\left[\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{3}}{4}\right)-\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{2}}{4}\right)\right]$ $=24 \times\left[\sin ^{-1} \frac{\sqrt{3}}{2}-\sin ^{-1} \frac{1}{\sqrt{2}}\right]$

$=24 \times\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$

$=24 \times \frac{\pi}{12}=2 \pi$