(4)

Let, $I={\int }_0^{\pi }\frac{(x+3) \sin x}{1+3 {\cos }^{2 }x}dx$          ... (i)

Replace x to $\pi$ – x,

$I={\int }_0^{\pi }\frac{(\pi –x+3) \sin (\pi –x)}{1+3 {\cos }^{2 }(\pi –x)}dx$

$\Rightarrow I={\int }_0^{\pi }\frac{(\pi +3) \sin x–x \sin x}{1+3 {\cos }^{2 }x}dx$          ... (ii)

Adding equation (i) and (ii), we get

$2I={\int }_0^{\pi }\frac{(x+6) \sin x}{1+3 {\cos }^{2 }x}dx$

Let cos x = t $\Rightarrow$ – sin x dx = dt

$2I=(\pi +6){\int }_1^{–1}\frac{–dt}{(1+3t^2)}=\left(\frac{\pi +6}{3}\right){\int }_{–1}^1\frac{dt}{t^2+{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2}$

$\Rightarrow 2I=\left(\frac{\pi +6}{3}\right)·\frac{1}{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}·{\tan }^{–1}\left(\frac{t}{1/\sqrt{3}}\right){|}_{–1}^1$

$\Rightarrow 2I=\left(\frac{\pi +6}{\sqrt{3}}\right)\left[{\tan }^{–1}\right(\sqrt{3})–{\tan }^{–1}(–\sqrt{3}\left)\right]$

$\Rightarrow 2I=\left(\frac{\pi +6}{\sqrt{3}}\right)·(\frac{\pi }{3}–(–\frac{\pi }{3}\left)\right)=(\frac{\pi +6}{\sqrt{3}}·\frac{2\pi }{3})$

$\Rightarrow I=\frac{\pi (\pi +6)}{3\sqrt{3}}$.

(3)

Let $I={\int }_{–1}^{3/2}$$\left|{\pi }^{2}x\sin \pi x\right|$$dx$

$={\pi }^2[{\int }_{–1}^{1}x\sin \pi xdx–{\int }_1^{3/2}x\sin \pi xdx]$

$={\pi }^2[2{\int }_0^{1}x\sin \pi xdx–{\int }_1^{3/2}x\sin \pi xdx]$

$={\pi }^2[2{[\frac{–x\cos \pi x}{\pi }+\frac{\sin \pi x}{{\pi }^2}]}_0^1–{[\frac{–x\cos \pi x}{\pi }+\frac{\sin \pi x}{{\pi }^2}]}_1^{3/2}]$

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

$=3\pi +1$.

(3)

$I_1={\int }_{–1/2}^{1}2xf\left(2x\right(1–2x\left)\right)dx$

Let $2x=t \Rightarrow 2dx=dt$

when $x=\frac{–1}{2}, t=–1$ and

When for x = 1, t = 2

$\therefore I_1=\frac{1}{2}{\int }_{–1}^{2}tf\left(t\right(1–t\left)\right)dt$

$\Rightarrow 2I_1={\int }_{–1}^2(1–t)f\left(\right(1–t\left)\right(1–(1–t)\left)\right)dt$

$\Rightarrow 2I_1={\int }_{–1}^{2}f\left(t\right(1–t\left)\right)dt–{\int }_{–1}^{2}tf\left(t\right(1–t\left)\right)dt$

$\Rightarrow 2I_1=I_2–2I_1 \Rightarrow 4I_1=I_2$

$\Rightarrow \frac{I_1}{I_2}=\frac{1}{4} i.e., \frac{I_2}{I_1}=4$

(3)

We have, $f\left(x\right)=7{\tan }^{8}x+7{\tan }^{6}x–3{\tan }^{4}x–3{\tan }^{2}x$

$=7{\tan }^{6}x({\tan }^{2}x+1)–3{\tan }^{2}x({\tan }^{2}x+1)$

$=(7{\tan }^{6}x–3{\tan }^{2}x){\sec }^{2}x$

$I_1={\int }_0^{\pi /4}(7{\tan }^{6}x–3{\tan }^{2}x){\sec }^{2}xdx$

Putting tan x = t $\Rightarrow {\sec }^{2}xdx=dt$

$I_1={\int }_0^1(7t^6–3t^2)dt={[t^7–t^3]}_0^1=0$

$I_2={\int }_0^{\pi /4}x(7{\tan }^{6}x–3{\tan }^{2}x){\sec }^{2}xdx$

$={\int }_0^1(7t^6–3t^2){\tan }^{–1}tdt$

$={\left[{\tan }^{–1}t\right(t^7–t^3\left)\right]}_0^1–{\int }_0^1\frac{t^7–t^3}{1+t^2}dt=0–{\int }_0^1\frac{t^3(t^4–1)}{1+t^2}dt$

$=–{\int }_0^1{t}^3(t^2–1)dt=–{\int }_0^1(t^5–t^3)dt=–{[\frac{t^6}{6}–\frac{t^4}{4}]}_0^1=\frac{1}{12}$

$\therefore 7I_1+12I_2=0+12\left(\frac{1}{12}\right)=1$.

(1)

From Leibnitz theorem,

$f\text{'}\left(x\right)=\left(\frac{x^4–8x^2+15}{e^{x^2}}\right)\left(2x\right)$

$=\frac{(x^2–3)(x^2–5)\left(2x\right)}{e^{x^2}}$

$=\frac{(x–\sqrt{3})(x+\sqrt{3})(x–\sqrt{5})(x+\sqrt{5})2x}{e^{x^2}}$

$\Rightarrow f\text{'}\left(x\right)=0 \Rightarrow x=\pm \sqrt{3},0,\pm \sqrt{5}$

Maxima at $x\in \{–\sqrt{3},\sqrt{3}\}$

Minima at $x\in \{–\sqrt{5},0,\sqrt{5}\}$

Hence, 2 points of maxima and 3 points of minima.

(2)

Let ${\log }_e$$x=t \Rightarrow \frac{1}{x}dx=dt$

$\Rightarrow I={\int }_2^4\frac{e^{{\displaystyle \frac{1}{t^2+1}}}}{e^{{\displaystyle \frac{1}{t^2+1}}}+e^{{\displaystyle \frac{1}{{(6–t)}^2+1}}}}dt$         ... (i)

$\Rightarrow I={\int }_2^4\frac{e^{{\displaystyle \frac{1}{{(6–t)}^2+1}}}}{e^{{\displaystyle \frac{1}{{(6–t)}^2+1}}}+e^{{\displaystyle \frac{1}{t^{2+1}}}}}dt$             $[\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(b+a–x\left)dx\right]$          ... (ii)

Adding (i) and (ii), we get

$2I={\int }_2^{4}dt={\left[t\right]}_2^4=4–2=2 \Rightarrow 2I=2 \Rightarrow I=1$.

(2)

We have, $I={\int }_0^{\pi /2}\frac{{\sin }^{{\displaystyle \frac{3}{2}}}x}{{\sin }^{{\displaystyle \frac{3}{2}}}x+{\cos }^{{\displaystyle \frac{3}{2}}}x}dx$          ... (i)

$I={\int }_0^{\pi /2}\frac{{\cos }^{{\displaystyle \frac{3}{2}}}x}{{\cos }^{{\displaystyle \frac{3}{2}}}x+{\sin }^{{\displaystyle \frac{3}{2}}}x}dx$          ... (ii)

Adding equation (i) and (ii), we get

$2I={\int }_0^{\pi /2}dx=\frac{\pi }{2} \Rightarrow I=\frac{\pi }{4}$

Let, $I_2={\int }_0^{2I}\frac{x\sin x \cos x}{{\sin }^{4}x+{\cos }^{4}x}dx={\int }_0^{\pi /2}\frac{x\sin x \cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$          ... (iii)

$\Rightarrow I_2={\int }_0^{\pi /2}\frac{({\displaystyle \frac{\pi }{2}}–x) \sin x \cos xdx}{{\cos }^{4}x+{\sin }^{4}x}$          (iv)

Adding (iii) and (iv), we get

$I_2=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{\tan x {\sec }^{2}xdx}{{\tan }^{4}x+1}$

Now, put ${\tan }^{2}x=t$

$I_2=\frac{\pi }{8}{\int }_0^{\infty }\frac{dt}{t^2+1}=\frac{\pi }{8}{\tan }^{–1}\left(t\right){|}_0^{\infty }=\frac{\pi }{8}·\frac{\pi }{2}=\frac{{\pi }^2}{16}$

(3)

$I(m,n)={\int }_0^1{x}^{m–1}{(1–x)}^{n–1}dx,m,n>0$

Let $x={\sin }^2\mathrm{\theta } \Rightarrow dx=2\mathrm{sin\theta } \mathrm{cos\theta } \mathrm{d\theta }$

$\therefore I(m,n)={\int }_0^{\pi /2}{\sin }^{2m–2}\mathrm{\theta }\times {(1–{\sin }^2\mathrm{\theta })}^{n–1}\times 2\mathrm{sin\theta cos\theta }d\mathrm{\theta }$

                         $=2{\int }_0^{\pi /2}{\sin }^{2m–1}\mathrm{\theta }\times {\cos }^{2n–1}\mathrm{\theta } d\mathrm{\theta }$

$\therefore I(9,14)+I(10,13)=2{\int }_0^{\mathrm{\pi }/2}{\sin }^{17}\mathrm{\theta } {\cos }^{27}\mathrm{\theta } \mathrm{d\theta }+2{\int }_0^{\mathrm{\pi }/2}{\sin }^{19}\mathrm{\theta } {\cos }^{25}\mathrm{\theta } \mathrm{d\theta }$

$=2{\int }_0^{\mathrm{\pi }/2}{\sin }^{17}\mathrm{\theta } {\cos }^{25}\mathrm{\theta }({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })d\mathrm{\theta }$

$=2{\int }_0^{\mathrm{\pi }/2}{\sin }^{17}\mathrm{\theta } {\cos }^{25}\mathrm{\theta } \mathrm{d\theta }=\mathrm{I}(9,13)$.

(3)

Let $I={\int }_{–\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{96x^2{\cos }^{2}x}{(1+e^x)}dx$             ... (i)

$\Rightarrow I={\int }_{–\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{96x^2{\cos }^{2}x}{(1+e^{–x})}dx$            ... (ii)

                       $[\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b–x\left)dx\right]$

Adding equations (i) and (ii), we get

    $2I={\int }_{–\pi /2}^{\pi /2}\frac{96x^2{\cos }^{2}x}{1+e^x}dx+{\int }_{–\pi /2}^{\pi /2}\frac{e^x\left(96x^2{\cos }^{2}x\right)}{e^x+1}dx$

                $=2{\int }_0^{\pi /2}96x^2{\cos }^{2}xdx$

$\Rightarrow I={\int }_0^{\pi /2}96x^2{\cos }^{2}xdx$

               $=48 {\int }_0^{\pi /2}{x}^2(1+\cos 2x)dx$

                $=48[{\left[\frac{x^3}{3}\right]}_0^{\pi /2}+{\int }_0^{\pi /2}{x}^2\cos 2xdx]$

$\Rightarrow I=48·\frac{{\pi }^3}{24}+{\left[\frac{x^2\sin 2x}{2}\right]}_0^{\pi /2}–{\int }_0^{\pi /2}x \sin 2x$

               $=48[\frac{{\pi }^3}{24}+{\left[\frac{x \cos 2x}{2}\right]}_0^{\pi /2}–\frac{1}{4}{[\sin 2x]}_0^{\pi /2}]$

$\Rightarrow I=48[\frac{{\pi }^3}{24}–\frac{\pi }{4}]=\pi (2{\pi }^2–12)$

So, $\alpha =2, \beta =–12$

$\therefore {(\alpha +\beta )}^2={(2–12)}^2={(–10)}^2=100$.

(4)

We have, $g\left(x\right)={\int }_0^{x}tf\left(t\right)dt$

$g\left(x\right)=x^2+x^{7/3} \Rightarrow g\text{'}\left(x\right)=2x+\frac{7}{3}{x}^{4/3}$

As, $f\left(x\right)=\frac{g\text{'}\left(x\right)}{x}=2+\frac{7}{3}{x}^{1/3} \Rightarrow f\left(r^3\right)=2+\frac{7r}{3}$

$\therefore \sum _{r=1}^{15}(2+\frac{7}{3}r)=30+\frac{7}{3}[1+2+...+15]=310$.