(1)

Ans.       1

Explanation :

Let        $I={\int }_{-\pi /4}^{\pi /4}\frac{dx}{1+\cos 2x}$

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

              $=\frac{1}{2}{\int }_{-\pi /4}^{\pi /4}{\sec }^{2}xdx={\int }_0^{\pi /4}{\sec }^{2}xdx$

              $={\left[\tan x\right]}_0^{\pi /4}=\tan \frac{\pi }{4}-\tan 0$

              $=1-0=1$

(4)

Ans.        $2(\sqrt{2}-1)$

Explanation :

Let          $I={\int }_0^{\pi /2}\sqrt{1-\sin 2x}dx$

               $={\int }_0^{\pi /2}\sqrt{{\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x}dx$

                $={\int }_0^{\pi /2}\sqrt{{(\cos x-\sin x)}^2}dx$

                $={\int }_0^{\pi /4}\sqrt{{(\cos x-\sin x)}^2}dx+{\int }_{\pi /4}^{\pi /2}\sqrt{{(\sin x-\cos x)}^2}dx$

                                                                 $[\because \cos x>\sin x\text{ for }x\in (0,\frac{\pi }{4}\left)\right]$

                                                                 $[\because \sin x>\cos x\text{ for }x\in (\frac{\pi }{4},\frac{\pi }{2}\left)\right]$

$={[\sin x+\cos x]}_0^{\pi /4}+{[-\cos x-\sin x]}_{\pi /4}^{\pi /2}$

$=[\sin \frac{{\displaystyle \pi }}{{\displaystyle 4}}+\cos \frac{{\displaystyle \pi }}{{\displaystyle 4}}-\sin 0-\cos 0]+[-\cos \frac{{\displaystyle \pi }}{{\displaystyle 2}}-\sin \frac{{\displaystyle \pi }}{{\displaystyle 2}}+\cos \frac{{\displaystyle \pi }}{{\displaystyle 4}}+\sin \frac{{\displaystyle \pi }}{{\displaystyle 4}}]$

$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1+(-0-1+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}})$

$=2\sqrt{2}-2=2(\sqrt{2}-1)$

(2)

Ans.       $e-1$

Explanation :

Let                    $I={\int }_0^{\pi /2}\cos x{e}^{\sin x}dx$

Put            $\sin x=t$

$\Rightarrow$ $\cos xdx=dt$

As $x\to 0$, then $t\to 0$,

and as $x\to \frac{\pi }{2}$, then $t\to 1$

$\therefore$          $I={\int }_0^1{e}^{t}dt$$={\left[e^t\right]}_0^1$

                $=e^1-e^0$

                $=e-1$

(1)

Ans.         $\log \frac{4}{3}$

Explanation:

Put   $\sin x=t$$\Rightarrow \cos xdx=dt,$      at $x=0,t=0$

                                                                at $x=\frac{\pi }{2},t=1$

So that the reduced integral is

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

                                $={\left[\log \right(1+t)-\log (2+t\left)\right]}_0^1$

                                $=[\log 2-\log 3]-[\log 1-\log 2]$

                                $=\log \frac{2}{3}-\log \frac{1}{2}=\log \frac{4}{3}$

(1)

Ans.         $\pi +2$

Explanation :

Put   $x=2\cos \theta$$\Rightarrow dx=-2\sin \theta d\theta$

At $x=0$$\Rightarrow 0=2\cos \theta$$\Rightarrow \cos \theta =\cos \frac{\pi }{2}$$\Rightarrow \theta =\frac{\pi }{2}$

and $x=2$$\Rightarrow 2=2\cos \theta$$\Rightarrow \cos \theta =\cos 0°$$\Rightarrow \theta =0$

Then,  ${\int }_0^2\sqrt{\frac{2+x}{2-x}}dx$

            $={\int }_{\pi /2}^0\sqrt{\frac{2+2\cos \theta }{2-2\cos \theta }}(-2\sin \theta )d\theta$

           $=-{\int }_0^{\pi /2}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}(-2\sin \theta )d\theta$

           $=2{\int }_0^{\pi /2}\sqrt{\frac{1+2{\cos }^2\frac{\theta }{2}-1}{1-(1-2{\sin }^2\frac{\theta }{2})}}.2\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta$

          $=2{\int }_0^{\pi /2}\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}} 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta$

           $=2{\int }_0^{\pi /2}2{\cos }^2\frac{\theta }{2}d\theta$

            $=2{\int }_0^{\pi /2}(1+\cos \theta )d\theta$

             $=2{[\theta +\sin \theta ]}_0^{\pi /2}$$=2[\frac{\pi }{2}+1]$

              $=\pi +2$

(3)

Ans.        20

Explanation :

${\int }_{-4}^4$$|x+2|$$dx={\int }_{-4}^{-2}-(x+2)dx+{\int }_{-2}^4(x+2)dx$

                          $={[\frac{{\displaystyle -x^2}}{{\displaystyle 2}}-2x]}_{-4}^{-2}+{[\frac{{\displaystyle x^2}}{{\displaystyle 2}}+2x]}_{-2}^4$

                          $=[\frac{{\displaystyle -4}}{{\displaystyle 2}}+4+\frac{{\displaystyle 16}}{{\displaystyle 2}}-8]+[\frac{{\displaystyle 16}}{{\displaystyle 2}}+8-\frac{{\displaystyle 4}}{{\displaystyle 2}}+4]$

                          $=2+18$

                          $=20$

(3)

Ans.        0

Explanation :

        ${\int }_0^{\pi /2}\frac{\cos x-\sin x}{1+\sin x\cos x}dx=I$               $\cdots \left(i\right)$

$\text{Now}$      $I={\int }_0^{\pi /2}\frac{\cos {\displaystyle (}{\displaystyle \frac{{\displaystyle \pi }}{{\displaystyle 2}}}{\displaystyle -}{\displaystyle x}{\displaystyle )}-\sin (\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dx$                 $[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]$

                  $={\int }_0^{\pi /2}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$      $\cdots \left(ii\right)$

$\text{On adding equation (i) and (ii),}$

                  $2I=0\Rightarrow I=0$

(1)

Ans.       0

Explanation :

           $I={\int }_{-1/2}^{1/2}\left(\cos x\right)\left[\log \right(\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}\left)\right]dx$          $\cdots \left(i\right)$

$\Rightarrow$ $I={\int }_{-1/2}^{1/2}\cos (-x)\left[\log \right(\frac{1+x}{1-x}\left)\right]dx$                   $[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]$

$\Rightarrow$$I=-{\int }_{-1/2}^{1/2}\cos x\left[\log \right(\frac{1-x}{1+x}\left)\right]dx$         $\cdots \left(ii\right)$

$\text{Adding (i) and (ii), we get}$

        $2I={\int }_{-1/2}^{1/2}\cos x\left[\log \right(\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}\left)\right]dx-{\int }_{-1/2}^{1/2}\cos x\left[\log \right(\frac{{\displaystyle 1-x}}{{\displaystyle 1+x}}\left)\right]dx$

$\text{or}$$2I=0$$\text{or}$$I=0$                    $[\because \log (1)=0]$

(2)

Ans.         $2-\sqrt{2}$

Explanation :

${\int }_0^{1.5}\left[x^2\right]dx={\int }_0^1\left[x^2\right]dx+{\int }_1^{\sqrt{2}}\left[x^2\right]dx+{\int }_{\sqrt{2}}^{1.5}\left[x^2\right]dx$

                       $=0+{\int }_1^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{1.5}2dx$

                       $={\left[x\right]}_1^{\sqrt{2}}+2{\left[x\right]}_{\sqrt{2}}^{1.5}$

                       $=\sqrt{2}-1+3-2\sqrt{2}$

                       $=2-\sqrt{2}$

(3)

Ans.         0

Explanation :

          $I={\int }_0^{\pi /2}\sin 2x\log \tan xdx$     $\cdots \left(1\right)$

$\Rightarrow$$I={\int }_0^{\pi /2}\sin 2(\frac{\pi }{2}-x)\log \tan (\frac{\mathrm{\pi }}{2}-\mathrm{x})dx$

                                          $[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]$

         $={\int }_0^{\pi /2}\sin 2x\log \cot xdx$         $\cdots \left(2\right)$

$\text{Adding (1) and (2), we get}$

$\Rightarrow$$2I={\int }_0^{\pi /2}\sin 2x[\log \tan x+\log \cot x]dx$

             $={\int }_0^{\pi /2}\sin 2x·\log [\tan x·\cot x]dx$

              $={\int }_0^{\pi /2}\sin 2x·\log 1dx$$[\because \tan x=\frac{1}{\cot x}]$

$\therefore$  $2I=0$                          $[\because \log 1=0]$

$\Rightarrow$$I=0$

(1)

Ans.         $\frac{4-\pi }{8}$

Explanation :

        ${\int }_0^{\pi /8}{\tan }^2\left(2x\right)dx={\int }_0^{\pi /8}\left({\sec }^2\right(2x)-1)dx$

                                         $={[\frac{{\displaystyle \tan \left(2x\right)}}{{\displaystyle 2}}-x]}_0^{\pi /8}$

                                         $=\frac{1}{2}-\frac{\pi }{8}+0-0=\frac{4-\pi }{8}$

(2)

Ans.     5

Explanation :

$I={\int }_{-1}^1\left(\right|x-2|+|x\left|\right)dx$

 $=-{\int }_{-1}^1(x-2)dx-{\int }_{-1}^{0}xdx+{\int }_0^{1}xdx$

 $=-{[\frac{{\displaystyle x^2}}{{\displaystyle 2}}-2x]}_{-1}^1-{\left[\frac{{\displaystyle x^2}}{{\displaystyle 2}}\right]}_{-1}^0+{\left[\frac{{\displaystyle x^2}}{{\displaystyle 2}}\right]}_0^1$

 $=1+\frac{3}{2}+\frac{5}{2}+\frac{1}{2}+\frac{1}{2}$

 $=\frac{10}{2}=5$

(4)

Ans.        $2(e-1)$

Explanation :

           $I={\int }_{-1}^1{e}^{\left|x\right|}dx$

             $={\int }_0^1{e}^{x}dx+{\int }_{-1}^0{e}^{-x}dx$

              $={\left[e^x\right]}_0^1-{\left[e^{-x}\right]}_{-1}^0$

              $=(e-1)-(1-e)$

              $=2e-2$

              $=2(e-1)$

(1)

$\text{By differentiating w.r.t. }x,\text{ we get }(1+x^2)f\left(x\right)=x\Rightarrow f\left(x\right)\text{ is odd}$

$\text{Required integral}={\int }_{-\pi /4}^{\pi /4}{\sec }^{2}xdx=2$

(3)

$\text{Let }x=\tan \theta$

${\int }_0^{\pi /4}\frac{\theta }{1+\tan \theta }{\sec }^2\theta d\theta$

$=\theta {\log }_e(1+\tan \theta )I_0^{\pi /4}-{\int }_0^{\pi /4}{\log }_e(1+\tan \theta )d\theta$$=\frac{\pi }{4}{\log }_{e}2-I$

$I={\int }_0^{\pi /4}{\log }_e(1+\tan \theta )d\theta$$...\left(1\right)$

$I={\int }_0^{\pi /4}{\log }_e(1+\tan (\frac{\pi }{4}-\theta \left)\right)d\theta$

$I={\int }_0^{\pi /4}{\log }_e\left(\frac{2}{1+\tan \theta }\right)d\theta$

$\text{By }\left(1\right)+\left(2\right)\Rightarrow 2I={\int }_0^{\pi /4}{\log }_{e}2d\theta =\frac{\pi }{4}{\log }_{e}2$

$\Rightarrow I=\frac{\pi }{8}{\log }_{e}2$

(1)

$\text{Given, }f\left(x\right)={\int }_0^{x{\tan }^{-1}x}\frac{e^{ t-\cos t}}{1+t^{2023}}dt$

$f\text{'}\left(x\right)=\frac{e^{ x{\tan }^{-1}x-\cos \left(x{\tan }^{-1}x\right)}}{1+{\left(x{\tan }^{-1}x\right)}^{2023}}.(\frac{x}{1+x^2}+{\tan }^{-1}x)$

$\text{For }x<0,{\tan }^{-1}x\in (-\frac{\pi }{2},0)$

$\text{For }x\ge 0,{\tan }^{-1}x\in [0,\frac{\pi }{2}]\Rightarrow x{\tan }^{-1}\forall x\in R$

$\text{And }\frac{x}{1+x^2}+{\tan }^{-1}x=$$\{\begin{array}{cc}>0& \text{For }x>0\\ <0& \text{For }x<0\\ 0& \text{For }x=0\end{array}$

$\text{So, }f\left(x\right)\text{ is minimum at }x=0$

$\text{Here minimum value is }f\left(0\right)={\int }_0^0=0.$

$\psi \left(x\right)\text{ and }\varphi \left(x\right)\text{ can be negative functions}$

(1)

$\Delta =0,a=5$

${\int }_0^{-10}f\left(x\right)dx={\int }_0^{5}f\left(x\right)dx+{\int }_{-5}^{10}f(x+5)dx+{\int }_{-85}^{10}f\left(x\right)dx=2{\int }_{-5}^{-10}dx=-10$

(4)

$2$$\left|I\right|$$=2{\int }_{\pi /8}^{3\pi /8}\frac{1+\cos 4x}{1-\cos 4x}dx$

$\Rightarrow$$\left|I\right|$$={\int }_{\pi /8}^{3\pi /8}\frac{12-(1-\cos 4x)}{1-\cos 4x}dx=12{\int }_{\pi /8}^{3\pi /8}\frac{1}{2{\sin }^{2}2x}dx-{\int }_{\pi /8}^{3\pi /8}dx=6{\int }_{\pi /8}^{3\pi /8}{\mathrm{cosec}}^{2}2xdx-\frac{\pi }{4}$

$={[-\frac{6}{2}\cot 2x]}_{\pi /8}^{3\pi /8}-\frac{\pi }{4}=-3\left[\right(1)+(1\left)\right]-\frac{\pi }{4}=-6-\frac{\pi }{4}$

(2)

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

(2)

${\int }_0^3\left(\frac{e^x+e^{–x}}{\left[x\right]!}\right)dx={\left\{\frac{e^x-e^{–x}}{1}\right\}}_0^1+{\left\{\frac{e^x-e^{–x}}{1}\right\}}_1^2+{\left\{\frac{e^x-e^{–x}}{2}\right\}}_2^3$

$=(e–\frac{1}{e})+(e^2–\frac{1}{e^2})–(e–\frac{1}{e})+\left(\frac{e^3–{\displaystyle \frac{1}{e^3}}}{2}\right)–\left(\frac{e^2–{\displaystyle e^{\frac{1}{2}}}}{2}\right)$

$=\frac{1}{2}\left\{\right(e^2–\frac{1}{e^2})+(e^3–\frac{1}{e^3}\left)\right\}$

(3)

Let

$I={\int }_0^{20\pi }({\sin }^{4}x+{\cos }^{4}x)dx$

    $=({\int }_0^{20\pi }{({\sin }^{2}x+{\cos }^{2}x)}^2–2{\sin }^{2}x{\cos }^{2}x)dx$

    $={\int }_0^{20\pi }(1–\frac{1}{2}{\sin }^2(2x)dx$

Period of ${\sin }^2\left(2x\right) \mathrm{is} \pi /2$

$\therefore I=40{\int }_0^{\pi /2}({\sin }^{4}x+{\cos }^{4}x)dx$

            $=40{\int }_0^{\pi /2}{\sin }^{4}xdx+40{\int }_0^{\pi /2}{\cos }^{4}xdx$

Now, ${\int }_0^{\pi /2}{\sin }^{4}xdx={\int }_0^{\pi /2}{\cos }^{4}xdx$

$\therefore I=40\times 2{\int }_0^{\pi /2}{\sin }^{4}xdx$

           $=80{\int }_0^{\pi /2}{\sin }^{4}xdx$

Using Walli's formula

${\int }_0^{\pi /2}{\sin }^{4}xdx=\frac{3·1}{4·2}\times \frac{\pi }{2}=\frac{3\pi }{16}$

$\therefore I=80\times \left(\frac{3\pi }{16}\right)=15\pi$

(3)

We have, $f\left(x\right)=f\text{'}\left(x\right)·f\text{'}\text{'}\left(x\right)$

From balancing degree,

        n = n – 1 + n – 2

$\Rightarrow$ n = 3

$\therefore$   f(x) is a cubic polynomial,

Also, f(0) = 0

Let say $f\left(x\right)=a{x}^3+b{x}^2+cx$

              $f\text{'}\left(x\right)=3a{x}^2+2bx+c$

              $f\text{'}\text{'}\left(x\right)=6ax+2b$

Now, $f\left(x\right)=f\text{'}\left(x\right)·f\text{'}\text{'}\left(x\right)$

$a{x}^3+b{x}^2+cx=(3a{x}^2+2bx+c)(6ax+2b)$

$\Rightarrow a{x}^3+b{x}^2+cx=18a^2{x}^3+(6ab+12ab)x^2+(4b^2+6ac)x+2bc$

Comparing coefficients, we get

        $a=18a^2$

$\Rightarrow a=\frac{1}{18}$        ($\because a\ne 0$)

         b = 18 ab

$\Rightarrow b=b$          $(\mathrm{Using} a=\frac{1}{18})$

       $c=4b^2+6ac$

       $c=4b^2+\frac{c}{3}$           $(\because a=\frac{1}{18})$

$\Rightarrow \frac{2c}{3}=4b^2 \Rightarrow c=6b^2$

        2bc = 0

$\Rightarrow b=0 \mathrm{or} c=0$

If $b=0 \Rightarrow c=0 \& \mathrm{if} c=0 \Rightarrow b=0$

$\therefore f\left(x\right)=\frac{1}{18}{x}^3 \Rightarrow f\text{'}\left(x\right)=\frac{x^2}{6}$

$\Rightarrow f\text{'}\text{'}\left(x\right)=\frac{x}{3}$

Now, $36(f\text{'}(2)+f\text{'}\text{'}(2)+{\int }_0^{2}f(x\left)dx\right)$

             $=36(\frac{4}{6}+\frac{2}{3}+{\int }_0^2\frac{x^3}{18}dx)$

             $=24+24+\frac{36}{18}\left(\frac{x^4}{4}\right){|}_0^2$

             $=48+\frac{2}{4}\left(2^4\right)=48+8=56$

(2)

We have, ${\int }_{–2}^2\left(\right|\sin x|+[x\sin x\left]\right)dx$

                    $={\int }_{–2}^2\left|\sin x\right|dx+{\int }_{–2}^2\left[x\sin x\right]dx$

                    $=2{\int }_0^2\sin xdx+2{\int }_0^2\left[x\sin x\right]dx$

                   $=2(–\cos x){|}_0^2+2({\int }_0^{\alpha }0dx+{\int }_{\alpha }^{2}dx)$          ($\because \alpha \text{ is such that }\alpha \text{ sin }\alpha =1$)

                    $=2(–\cos 2+1)+2(2–\alpha )$

                    $=6–2\cos 2–2\alpha$

On comparing, we get $\beta =–2\alpha$

Now, $\beta \sin \left(\frac{\beta }{2}\right)=(–2\alpha )\sin \left(\frac{–2\alpha }{2}\right)$

$=2\alpha \sin \alpha =2\times 1=2$

(3)