Let,

$I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin 2x {\cos }^{\frac{11}{2}}x{\left(1+{\cos }^{\frac{5}{2}}x\right)}^{\frac{1}{2}}dx$

Let $\cos x=t^2\Rightarrow \sin xdx=-2t dt$

$\therefore I=-4\underset{1}{\overset{0}{\int }}{t}^2\cdot t^{11}\sqrt{\left(1+t^5\right)}\left(t\right)dt$

$\Rightarrow I=4\underset{0}{\overset{1}{\int }}{t}^{14}\sqrt{\left(1+t^5\right)}dt$

Now, let $1+t^5=k^2$

$\Rightarrow 5t^{4}dt=2k dk$

$\therefore I=4\cdot \underset{1}{\overset{\sqrt{2}}{\int }}{\left(k^2-1\right)}^2\cdot k·\frac{2k}{5}dk$

$\Rightarrow I=\frac{8}{5}\underset{1}{\overset{\sqrt{2}}{\int }}{k}^6-2k^4+k^2 dk$

$\Rightarrow I=\frac{8}{5}{\left[\frac{k^7}{7}-\frac{2k^5}{5}+\frac{k^3}{3}\right]}_1^{\sqrt{2}}$

$\Rightarrow I=\frac{8}{5}\left[\frac{8\sqrt{2}}{7}-\frac{8\sqrt{2}}{5}+\frac{2\sqrt{2}}{3}-\frac{1}{7}+\frac{2}{5}-\frac{1}{3}\right]$

$\Rightarrow I=\frac{8}{5}\left[\frac{22\sqrt{2}}{105}-\frac{8}{105}\right]$

$\Rightarrow 525\cdot I=176\sqrt{2}-64$

Hence, on comparing we get, $n=176$