(1)

$I={\int }_0^{\pi /4}f\left(\sin 2x\right)\sin x dx+{\int }_{\pi /4}^{\pi /2}f\left(\sin 2x\right)\sin x dx+\alpha {\int }_0^{\pi /4}f\left(\cos 2x\right)\cos x dx=0$ 

$\text{Apply }{\int }_0^{a}f\left(x\right) dx={\int }_0^{a}f(a-x) dx\text{ in the first part and put } x-\frac{\pi }{4}=t\text{ in the second part, we get}$

$I={\int }_0^{\pi /4}f\left(\cos 2x\right)\sin (\frac{\pi }{4}-x)dx+{\int }_0^{\pi /4}f\left(\cos 2t\right)\sin (\frac{\pi }{4}+t)dt+\alpha {\int }_0^{\pi /4}f\left(\cos 2x\right)\cos x dx=0$

$={\int }_0^{\pi /4}f\left(\cos 2x\right)[2\sin \frac{\pi }{4}·\cos x+\alpha \cos x]dx=0$

$(\alpha +\sqrt{2}){\int }_0^{\pi /4}f\left(\cos 2x\right)\cos x dx=0 \therefore \alpha =-\sqrt{2}$

$[\because \text{ In }(0,\frac{\pi }{4}), f(\cos \left(2x\right)\left)\text{ and }\cos x\text{ are not zero.}\right]$

(2)

$f\left(x\right)=$$\{\begin{array}{cc}{e}^{\min \{x^2, x-[x\left]\right\}},& x\in [0,1)\\ e^{[x-{\log }_{e}x]},& x\in [1,2)\end{array}$

$\text{For }x\in [0,1);$$\min \{x^2, \{x\left\}\right\}=x^2$ 

and for $x\in$[1,2); $[x-{\log }_{e}x]=1$ 

$\therefore$$f\left(x\right)=$$\{\begin{array}{cc}{e}^{x^2},& x\in [0,1)\\ e,& x\in [1,2)\end{array}$

$\text{Now, }{\int }_0^{2}xf\left(x\right)={\int }_0^{1}x{e}^{x^2} dx+{\int }_1^{2}xedx$

$=\frac{1}{2}(e-1)+\frac{1}{2}(4-1)e=2e-\frac{1}{2}$

(4)

Put $e^x=t\Rightarrow e^x dx=dt$ and on solving, we get ${\log }_e\left(\frac{32}{27}\right)$

(4)

$\frac{e^{-\pi /4}+{\int }_0^{\pi /4}{e}^{-x}{\tan }^{50}x dx}{{\displaystyle {\int }_0^{\pi /4}{e}^{-x}({\tan }^{49}x+{\tan }^{51}x)dx}}$

First, simplify, ${\int }_0^{\pi /4}{e}^{-x}{\tan }^{50}xdx$

$={[-e^{-x}{\left(\tan x\right)}^{50}]}_0^{\pi /4}+{\int }_0^{\pi /4}{e}^{-x}\left(50\right){\tan }^{49}x·{\sec }^{2}x dx$

$=-e^{-\pi /4}+0+50{\int }_0^{\pi /4}{e}^{-x}{\left(\tan x\right)}^{49}(1+{\tan }^{2}x) dx$

$=-e^{-\pi /4}+50{\int }_0^{\pi /4}{e}^{-x}[{\left(\tan x\right)}^{51}+{\left(\tan x\right)}^{49}]dx$

Now, $\frac{-e^{-\pi /4}+{\int }_0^{\pi /4}{e}^{-x}{\left(\tan x\right)}^{50}dx}{{\int }_0^{\pi /4}{e}^{-x}({\tan }^{49}x+{\tan }^{51}x)dx}$

$=\frac{50{\int }_0^{\pi /4}{e}^{-x}({\tan }^{51}x+{\tan }^{49}x)dx}{{\int }_0^{\pi /4}{e}^{-x}({\tan }^{49}x+{\tan }^{51}x)dx}=50$

$\therefore$ $\frac{e^{-\pi /4}+{\displaystyle {\int }_0^{\pi /4}{e}^{-x}{\tan }^{50}x dx}}{{\displaystyle {\int }_0^{\pi /4}{e}^{-x}({\tan }^{49}x+{\tan }^{51}x)dx}}=50$

(3)

Let $I={\int }_0^1\frac{1}{(5+2x-2x^2)(1+e^{(2-4x)})}dx$

$I={\int }_0^1\frac{1}{(5+2x(1-x\left)\right)}·\frac{e^{4x}}{(e^{4x}+e^2)} dx$                  ...(i)

We know that, ${\int }_a^{b}f\left(x\right) dx={\int }_a^{b}f(b+a-x)dx$

$\therefore$   $I={\int }_0^1\frac{1}{5+2(1-x)·x}·\frac{e^{4(1-x)}}{e^{4(1-x)}+e^2} dx$            ...(ii)

Adding (i) and (ii), we get

$\Rightarrow 2I={\int }_0^1\frac{1}{5+2(1-x)·x}\left[\frac{{\displaystyle e^{4x}{e}^{4(1-x)}+e^{4x}{e}^2+e^{4x}{e}^{4(1-x)}+e^{4(1-x)}·e^2}}{{\displaystyle (e^{4(1-x)}+e^2)(e^{4x}+e^2)}}\right]dx$

$\Rightarrow 2I={\int }_0^1\frac{dx}{5+2(1-x)·x}={\int }_0^1\frac{dx}{\frac{11}{2}-2{(x-\frac{1}{2})}^2}$

$={\int }_0^1\frac{dx}{(\frac{11}{4}-{(x-\frac{1}{2})}^2)}$

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

$\therefore$  $\alpha =\sqrt{11}\text{ and }\beta =\sqrt{10} \therefore {\alpha }^4-{\beta }^4=121-100=21$

(1)

$\underset{n\to \infty }{\lim }[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\dots +\frac{1}{2n}]$

$=\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{r+n}=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=1}^n\frac{1}{1+\frac{r}{n}}={\int }_0^1\frac{dx}{1+x}=\ln [1+x]{|}_0^1=\ln 2$

(4)

$I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{x+\frac{\pi }{4}}{2-\cos 2x} dx \text{...(i)}$

Replace $x$ by $-x$, we get

$I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{-x+\frac{\pi }{4}}{2-\cos 2x} dx \text{...(ii)}$

Adding (i) and (ii), we get

$2I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\pi /2}{2-\cos 2x}dx\Rightarrow I=\frac{\pi }{4}·2{\int }_0^{\frac{\pi }{4}}\frac{dx}{2-\cos 2x}dx$

$\Rightarrow I=\frac{\pi }{4}·2{\int }_0^{\frac{\pi }{4}} \frac{(1+{\tan }^{2}x) dx}{2(1+{\tan }^{2}x)-(1-{\tan }^{2}x)}$

$=\frac{\pi }{2}{\int }_0^{\frac{\pi }{4}} \frac{{\sec }^{2}x dx}{2(1+{\tan }^{2}x)-(1-{\tan }^{2}x)}$

Let $\tan x=t\Rightarrow {\sec }^{2}x dx=dt$

at $x=0,t=0$ and $x=\frac{\pi }{4},t=1$

$I=\frac{\pi }{2}{\int }_0^1\frac{dt}{3t^2+1}\Rightarrow I=\frac{\pi }{6}{\int }_0^1\frac{dt}{t^2+{\left(\frac{1}{\sqrt{3}}\right)}^2}=\frac{\pi }{6}\sqrt{3}·{\left[{\tan }^{-1}\sqrt{3}t\right]}_0^1$

$\Rightarrow I=\frac{\pi }{2\sqrt{3}}{\tan }^{-1}\sqrt{3} I=\frac{{\pi }^2}{6\sqrt{3}}$

(1)

(2)

Case 1: When $x<0$

$f\left(x\right)={\int }_0^2{e}^{t-x}dt=e^{-x}{\left(e^t\right)}_0^2=e^{-x}(e^2-1)$

Case 2: When $0<x<2$

$f\left(x\right)={\int }_0^x{e}^{x-t}dt+{\int }_x^2{e}^{t-x}dt \Rightarrow f\left(x\right)=e^x{(-e^{-t})}_0^x+e^{-x}{\left(e^t\right)}_x^2$

$=-e^x(e^{-x}-1)+e^{-x}(e^2-e^x)=-1+e^x+e^2{e}^{-x}-1=e^x+e^{-x}{e}^2-2$

Case 3: When $x\ge 2$

$f\left(x\right)={\int }_0^2{e}^{(x-t)}dt=e^x{(-e^{-t})}_0^2=-e^x(e^{-2}-1)$

Hence,

$f\left(x\right)=$$\{\begin{array}{cc}{e}^{-x}(e^2-1),& x<0 \\ e^x+e^{-x}{e}^2-2,& 0\le x<2\\ (1-e^{-2})e^x,& x\ge 2\end{array}$

$f\left(x\right)$ is decreasing for $x<0$ and increasing for $x\ge 2$.  

$f\left(x\right)$ is also continuous.

For $0\le x\le 2$, $f\left(x\right)$ is minimum at $x=1$.  

(Using A.M.–G.M. inequality)

$\text{Hence, the minimum value }f\left(x\right)=f\left(1\right)=2(e-1)$

(1)

Let $I=16{\int }_1^2\frac{dx}{x^3{(x^2+2)}^2}=16{\int }_1^2\frac{1}{x^7{(1+\frac{2}{x^2})}^2}dx$

Let $1+\frac{2}{x^2}=p \Rightarrow -\frac{4}{x^3}dx=dp \Rightarrow \frac{dx}{x^3}=\frac{-1}{4}dp$

$I=-4{\int }_3^{3/2}\frac{dp}{{\displaystyle \frac{4}{{(p-1)}^2}}{\displaystyle ·}{p}^2}$

$=-4{\int }_3^{3/2}\frac{{(p-1)}^{2}dp}{4p^2} =-{\int }_3^{3/2}\frac{p^2+1-2p}{p^2}dp$

$=-{\int }_3^{3/2}(1+\frac{{\displaystyle 1}}{{\displaystyle p^2}}-\frac{{\displaystyle 2}}{{\displaystyle p}})dp=-{[p-\frac{{\displaystyle 1}}{{\displaystyle p}}-2\log p]}_3^{3/2}$

$=-\left[\right\{\frac{{\displaystyle 5}}{{\displaystyle 6}}-2{\log }_e\left(\frac{{\displaystyle 3}}{{\displaystyle 2}}\right)\}-\{\frac{{\displaystyle 8}}{{\displaystyle 3}}-2{\log }_{e}3\left\}\right]$

$=-[2{\log }_{e}2-\frac{{\displaystyle 11}}{{\displaystyle 6}}]=\frac{11}{6}-{\log }_e{2}^2=\frac{11}{6}-{\log }_{e}4$