(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}$

[IMAGE 163]

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

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

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

(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}$

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