(4)

Let $L=\underset{x\to \pi /2}{\lim }\left[\frac{{\int }_{x^3}^{{(\pi /2)}^3}\left(\sin \right(2t^{1/3})+\cos (t^{1/3}\left)\right) dt}{{(x-\frac{\pi }{2})}^2}\right]$ $\frac{0}{0}\text{ form, so by L'Hospital rule}$

$=\underset{x\to \frac{\pi }{2}}{\lim }\frac{{\displaystyle \frac{d}{dx}}{\int }_{x^3}^{{(\pi /2)}^3}\left(\sin \right(2t^{1/3})+\cos (t^{1/3}\left)\right)dt}{2(x-\frac{\pi }{2})}$ $\sin (2\times \frac{\pi }{2})·\frac{d}{dx}{\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)}^3-\sin \left(2x\right)·\frac{d}{dx}{x}^3$

$=\underset{x\to \pi /2}{\lim }\frac{{\displaystyle +(\cos \frac{{\displaystyle \pi }}{{\displaystyle 2}}\frac{{\displaystyle d}}{{\displaystyle dx}}{\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)}^3-\cos x·\frac{{\displaystyle d}}{{\displaystyle dx}}{x}^3)}}{2(x-\frac{\pi }{2})}$    (By Leibnitz rule of integration)

$=\underset{x\to \pi /2}{\lim }\frac{-3x^2\sin 2x-3x^2\cos x}{2(x-\frac{\pi }{2})} \left(\frac{{\displaystyle 0}}{{\displaystyle 0}}\text{ form}\right)$

$=\underset{x\to \pi /2}{\lim }\frac{-6x\sin 2x-6x^2\cos 2x-6x\cos x+3x^2\sin x}{2}$

$=\frac{{\displaystyle 6}\frac{{\pi }^2}{4}+\frac{3{\pi }^2}{4}}{2}=\frac{9{\pi }^2}{8}$

(2)

$I={\int }_{1/4}^{3/4}\cos \left(2{\cot }^{-1}\sqrt{\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}}\right)dx$

$={\int }_{1/4}^{3/4}\cos \left(2{\tan }^{-1}\sqrt{\frac{{\displaystyle 1+x}}{{\displaystyle 1-x}}}\right)dx$

$={\int }_{1/4}^{3/4} \frac{1-{\tan }^2\left({\tan }^{-1}\sqrt{\frac{1+x}{1-x}}\right)}{ 1+{\tan }^2\left({\tan }^{-1}\sqrt{\frac{1+x}{1-x}}\right)}dx$

$={\int }_{1/4}^{3/4} \frac{{\displaystyle 1}{\displaystyle -}\frac{1+x}{1-x}}{{\displaystyle 1}{\displaystyle +}\frac{1+x}{1-x}}dx$

$={\int }_{1/4}^{3/4}\frac{-2x}{2} dx={(-\frac{{\displaystyle x^2}}{{\displaystyle 2}})}_{1/4}^{3/4}=\frac{-9}{32}+\frac{1}{32}=\frac{-8}{32}=\frac{-1}{4}$

(4)

Let $I={\int }_0^{\pi /4}\frac{xdx}{{\sin }^4\left(2x\right)+{\cos }^4\left(2x\right)}$

Put $2x=t$

$\Rightarrow 2dx=dt\Rightarrow dx=\frac{dt}{2}$    $\therefore I={\int }_0^{\pi /2}\frac{t/2·dt/2}{{\sin }^{4}t+{\cos }^{4}t}$

$\Rightarrow I=\frac{1}{4}{\int }_0^{\pi /2}\frac{t dt}{{\sin }^{4}t+{\cos }^{4}t}$                                                 ...(i)

$\Rightarrow I=\frac{1}{4}{\int }_0^{\pi /2}\frac{(\frac{\pi }{2}-t) dt}{{\sin }^{4}t+{\cos }^{4}t}$                                                 ...(ii)

Adding (i) and (ii), we get

$2I=\frac{\pi }{8}{\int }_0^{\pi /2}\frac{dt}{{\sin }^{4}t+{\cos }^{4}t}\Rightarrow 2I=\frac{\pi }{8}{\int }_0^{\pi /2}\frac{{\sec }^{4}t dt}{{\tan }^{4}t+1}$

Let $\tan t=y\Rightarrow {\sec }^{2}t dt=dy$

$\therefore 2I=\frac{\pi }{8}{\int }_0^{\infty }\frac{(1+y^2) dy}{1+y^4}$

$\Rightarrow 2I=\frac{\pi }{8}{\int }_0^{\infty }\frac{(\frac{1}{y^2}+1) dy}{y^2+\frac{1}{y^2}-2+2}=\frac{\pi }{8}{\int }_0^{\infty }\frac{(\frac{1}{y^2}+1) dy}{{{\displaystyle (}{\displaystyle y}{\displaystyle -}{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle y}}}{\displaystyle )}}^2+2}$

Let $y-\frac{1}{y}=u\Rightarrow (1+\frac{1}{y^2})dy=du,$ we get $2I=\frac{\pi }{8}{\int }_{-\infty }^{\infty }\frac{du}{u^2+2}$

$\Rightarrow 2I=\frac{\pi }{8}{\left[\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}{\tan }^{-1}\right(\frac{{\displaystyle u}}{{\displaystyle \sqrt{2}}}\left)\right]}_{-\infty }^{\infty }$

$\Rightarrow 2I=\frac{\pi }{8}\left[\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}{\tan }^{-1}\right(\infty )-\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}{\tan }^{-1}(-\infty \left)\right]$

$\Rightarrow I=\frac{\pi }{16\sqrt{2}}(\frac{{\displaystyle \pi }}{{\displaystyle 2}}+\frac{{\displaystyle \pi }}{{\displaystyle 2}})=\frac{{\pi }^2}{16\sqrt{2}}=\frac{\sqrt{2}{\pi }^2}{32}$

(1)

Let  $I={\int }_0^{\pi /3}{\cos }^{4}x dx={\int }_0^{\pi /3}{\left(\frac{{\displaystyle 1+\cos 2x}}{{\displaystyle 2}}\right)}^2 dx$

$=\frac{1}{4}{\int }_0^{\pi /3}(1+{\cos }^{2}2x+2\cos 2x) dx$

$=\frac{1}{4}{\int }_0^{\pi /3}(1+2\cos 2x+(\frac{{\displaystyle 1+\cos 4x}}{{\displaystyle 2}}\left)\right) dx$

$=\frac{1}{4}{\int }_0^{\pi /3}(\frac{{\displaystyle 3}}{{\displaystyle 2}}+2\cos 2x+\frac{{\displaystyle \cos 4x}}{{\displaystyle 2}}) dx$

$=\frac{1}{4}{[\frac{{\displaystyle 3}}{{\displaystyle 2}}x+\sin 2x+\frac{{\displaystyle \sin 4x}}{{\displaystyle 8}}]}_0^{\pi /3}$

$=\frac{1}{4}[\frac{{\displaystyle \pi }}{{\displaystyle 2}}+\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}-\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 16}}]=\frac{\pi }{8}+\frac{7\sqrt{3}}{64}\Rightarrow a=\frac{1}{8}, b=\frac{7}{64}$

$\therefore 9a+8b=\frac{9}{8}+\frac{7}{8}=2$

(3)

Let $I={\int }_0^1{(2x^3-3x^2-x+1)}^{1/3}dx$

$\Rightarrow I={\int }_0^1{[2{(1-x)}^3-3{(1-x)}^2-(1-x)+1]}^{1/3}dx$

$\Rightarrow I={\int }_0^1(-2x^3+3x^2+x-1)dx$

$\Rightarrow I={\int }_0^1-(2x^3-3x^2-x+1)dx$

$\Rightarrow I=-I\Rightarrow 2I=0\Rightarrow I=0$

(4)

${\int }_0^1\frac{1}{\sqrt{3+x}+\sqrt{1+x}} dx$

$={\int }_0^1\frac{1}{\sqrt{3+x}+\sqrt{1+x}}\times \frac{\sqrt{3+x}-\sqrt{1+x}}{\sqrt{3+x}-\sqrt{1+x}} dx$        [$\because$Rationalising the denominator]

$={\int }_0^1\frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} dx=\frac{1}{2}{\int }_0^1(\sqrt{3+x}-\sqrt{1+x}) dx$

$=\frac{1}{2}[$$\frac{{\displaystyle 2{(3+x)}^{3/2}}}{{\displaystyle 3}}$${|}_0^1$$-$$\frac{{\displaystyle 2}}{{\displaystyle 3}}{(1+x)}^{3/2}$${|}_0^1$$]$

$=3-\frac{2\sqrt{2}}{3}-\sqrt{3}=a+b\sqrt{2}+c\sqrt{3}\Rightarrow a=3, b=\frac{-2}{3}, c=-1$

$\therefore 2a+3b-4c=2\times 3+3\times \left(\frac{{\displaystyle -2}}{{\displaystyle 3}}\right)-4(-1)=8$

(1)

Let $I={\int }_0^{\pi }\frac{dx}{1+a^2-2a\cos x}$

$={\int }_0^{\pi }\frac{dx}{1+a^2-2a\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}$

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

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

$I=2{\int }_0^{\infty }\frac{dt}{(1+a^2)(1+t^2)-2a+2a{t}^2}$

$=2{\int }_0^{\infty }\frac{dt}{1+a^2+t^2+a^2{t}^2-2a+2a{t}^2}$

$=2{\int }_0^{\infty }\frac{dt}{(1+a^2-2a)+(t^2+a^2{t}^2+2a{t}^2)}$

$I=2{\int }_0^{\infty }\frac{dt}{{(1-a)}^2+{(t+at)}^2}$

Let $t+at=u \Rightarrow dt+a dt=du$

$\Rightarrow I=\frac{2}{a+1}{\int }_0^{\infty }\frac{du}{{(1-a)}^2+u^2}=\frac{2}{(a+1)}\times \frac{1}{1-a}\times {\left[{\tan }^{-1}\right(\frac{{\displaystyle u}}{{\displaystyle 1-a}}\left)\right]}_0^{\infty }$

$=\frac{2}{1-a^2}\times \frac{\pi }{2}=\frac{\pi }{1-a^2}$

(3)

Let $I={\int }_{-\pi /2}^{\pi /2}(\frac{{\displaystyle x^2\cos x}}{{\displaystyle 1+{\pi }^x}}+\frac{{\displaystyle 1+{\sin }^{2}x}}{{\displaystyle 1+e^{{\left(\sin x\right)}^{2023}}}})dx$                  ...(i)

$\Rightarrow I={\int }_{-\pi /2}^{\pi /2}\frac{x^2\cos x}{1+{\pi }^{-x}} +\frac{1+{\sin }^{2}x}{1+e^{-\sin x^{2023}}} dx$                 ...(ii)

Adding (i) and (ii), we get

$2I={\int }_{-\pi /2}^{\pi /2}(x^2\cos x+1+{\sin }^{2}x)dx$

$\Rightarrow I={\int }_0^{\pi /2}(x^2\cos x+1+{\sin }^{2}x)dx$

$={\int }_0^{\pi /2}{x}^2\cos x dx+{\int }_0^{\pi /2}(1+{\sin }^{2}x)dx$                     ...(iii)

Let $I_1={\int }_0^{\pi /2}{x}^2\cos x dx={\left[x^2\right(\sin x)-\int 2x\sin x dx]}_0^{\pi /2}$

           $={[x^2\sin x+2x\cos x-2\sin x]}_0^{\pi /2}$

$I_1={\left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)}^2\sin \frac{\pi }{2}+2\frac{\pi }{2}\cos \frac{\pi }{2}-2\sin \frac{\pi }{2}-0=\frac{{\pi }^2}{4}-2$

And $I_2={\int }_0^{\pi /2}(1+{\sin }^{2}x)dx={\int }_0^{\pi /2}[1+(\frac{{\displaystyle 1-\cos 2x}}{{\displaystyle 2}}\left)\right]dx$

$={[\frac{{\displaystyle 3}}{{\displaystyle 2}}x-\frac{{\displaystyle 1}}{{\displaystyle 4}}\sin 2x]}_0^{\pi /2}=\frac{3}{2}\times \frac{\pi }{2}-\frac{1}{4}\sin 2\times \frac{\pi }{2}=\frac{3\pi }{4}$

From (iii), $I=\frac{{\pi }^2}{4}-2+\frac{3\pi }{4}=\frac{\pi }{4}(\pi +3)-2$

Comparing with $\frac{\pi }{4}(\pi +a)-2,\text{ we get }a=3$

(2)

Let $I=\underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{1}{{(x-\frac{\pi }{2})}^2}{\int }_{x^3}^{{\left(\frac{\pi }{2}\right)}^3}\cos \right(t^{1/3}\left)dt\right)$ $\left(\frac{0}{0}\mathrm{form}\right)$

By Leibnitz theorem,

$\frac{d}{dx}{\int }_{h\left(x\right)}^{g\left(x\right)}f\left(t\right) dt=f\left(g\right(x\left)\right)\times g\text{'}\left(x\right)-f\left(h\right(x\left)\right)\times h\text{'}\left(x\right)$

$\therefore$    By applying L'Hospital rule, we get

$I=\underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{-\cos x\left(3x^2\right)}{2(x-\frac{\pi }{2})}\right)$

$=\underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{3x^2\sin x-6x\cos x}{2}\right) \text{(Applying L'Hospital rule)}$

$=\frac{3{\pi }^2}{8}$

(1)

We have, $\underset{x\to 0}{\lim }\frac{x{\int }_0^{x}f\left(t\right) }{e^{x^2}-1}dt=\alpha$

Using L'Hospital's rule, we get $\underset{x\to 0}{\lim }\frac{xf\left(x\right)+{\int }_0^{x}f\left(t\right) dt}{2x{e}^{x^2}}=\alpha$

Again applying L'Hospital's rule, we get

       $\underset{x\to 0}{\lim }\frac{f\left(x\right)+xf\text{'}\left(x\right)+f\left(x\right)}{2e^{x^2}+4x^2{e}^{x^2}}=\alpha \Rightarrow \alpha =\frac{2f\left(0\right)}{2}=f\left(0\right)$

$\Rightarrow \alpha =\frac{1}{2} [\because f(0)=\frac{{\displaystyle 1}}{{\displaystyle 2}}]$

$\therefore 8{\alpha }^2=8\times \frac{1}{4}=2$