(176)

Let $I=525{\int }_0^{\frac{\pi }{2}}\sin 2x·{\cos }^{\frac{11}{2}}x{(1+{\cos }^{\frac{{\displaystyle 5}}{{\displaystyle 2}}}x)}^{\frac{1}{2}}dx$

Put $\cos x=t^2$

       $-\sin x dx=2 t dt$

$I=-525{\int }_1^{0}2\sin x·t^2·t^{11}{(1+t^5)}^{\frac{1}{2}}·\frac{2t dt}{\sin x}$

$=525\times 4{\int }_0^1{t}^3·t^{11}{(1+t^5)}^{\frac{1}{2}} dt$

$I=2100{\int }_0^1{t}^{14}\sqrt{(1+t^5)} dt$

Let $1+t^5=k^2\Rightarrow 5t^4 dt=2k dk\Rightarrow t^4 dt=\frac{2k dk}{5}$

$\therefore I=2100{\int }_1^{\sqrt{2}}{(k^2-1)}^2·k·\frac{2k dk}{5}$

         $=\frac{2100\times 2}{5}{\int }_1^{\sqrt{2}}{k}^2{(k^2-1)}^2 dk$

        $=420\times 2{\int }_1^{\sqrt{2}}(k^6-2k^4+k^2) dk$

        $=840{[\frac{{\displaystyle k^7}}{{\displaystyle 7}}-\frac{{\displaystyle 2k^5}}{{\displaystyle 5}}+\frac{{\displaystyle k^3}}{{\displaystyle 3}}]}_1^{\sqrt{2}}$

         $=840[\frac{{\displaystyle 8\sqrt{2}}}{{\displaystyle 7}}-\frac{{\displaystyle 8\sqrt{2}}}{{\displaystyle 5}}+\frac{{\displaystyle 2\sqrt{2}}}{{\displaystyle 3}}-(\frac{{\displaystyle 1}}{{\displaystyle 7}}-\frac{{\displaystyle 2}}{{\displaystyle 5}}+\frac{{\displaystyle 1}}{{\displaystyle 3}}\left)\right]$

          $=840\left[\frac{{\displaystyle 22\sqrt{2}-8}}{{\displaystyle 105}}\right]=8[22\sqrt{2}-8]=176\sqrt{2}-64$ 

So, $n=176$

(15)

 $\text{Let }I={\int }_0^{\pi }\frac{x^2\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x} dx$

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

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

$=2\pi {\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x} dx-{\pi }^2{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x} dx$

$=2\pi ·\frac{\pi }{4}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x} -{\pi }^2{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x} dx$

$=\frac{-{\pi }^2}{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x} =\frac{-{\pi }^2}{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{1-2{\sin }^{2}x+{\cos }^{2}x} dx$

$=\frac{-{\pi }^2}{2}{\int }_0^{\pi /2}\frac{\sin 2x}{1+{\cos }^{2}2x} dx$

$\text{Put }\cos 2x=t \therefore I=\frac{-{\pi }^2}{2}{\int }_1^{-1}\frac{-dt}{2(1+t^2)}$

$=\frac{-{\pi }^2}{2}{\int }_0^1\frac{dt}{1+t^2}=\frac{-{\pi }^2}{2}{\left({\tan }^{-1}t\right)}_0^1=\frac{-{\pi }^2}{2}\left(\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)=\frac{-{\pi }^3}{8}$

$\therefore$  $|\frac{120}{{\pi }^3}{\int }_0^{\pi }\frac{x^2\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x} |$$=\frac{120}{{\pi }^3}\times \frac{{\pi }^3}{8}=15$