If the integral $$525 \int_\limits0^{\frac{\pi}{2}} \sin 2 x \cos ^{\frac{11}{2}} x\left(1+\operatorname{Cos}^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x$$ is equal to $(n \sqrt{2}-64)$, then $n$ is equal to _________.

<p>$$I=\int_\limits0^{\frac{\pi}{2}} \sin 2 x \cdot(\cos x)^{\frac{11}{2}}\left(1+(\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} d x$$</p> <p>Put $\cos x=t^2 \Rightarrow \sin x d x=-2 t d t$</p> <p>$$\begin{aligned} & \therefore \mathrm{I}=4 \int_\limits0^1 \mathrm{t}^2 \cdot \mathrm{t}^{11} \sqrt{\left(1+\mathrm{t}^5\right)}(\mathrm{t}) \mathrm{dt} \\ & \mathrm{I}=4 \int_\limits0^1 \mathrm{t}^{14} \sqrt{1+\mathrm{t}^5} \mathrm{dt} \end{aligned}$$</p> <p>Put $1+\mathrm{t}^5=\mathrm{k}^2$</p> <p>$\Rightarrow 5 \mathrm{t}^4 \mathrm{dt}=2 \mathrm{k} \mathrm{dk}$</p> <p>$$\therefore \mathrm{I}=4 \cdot \int_\limits1^{\sqrt{2}}\left(\mathrm{k}^2-1\right)^2 \cdot \mathrm{k} \frac{2 \mathrm{k}}{5} \mathrm{dk}$$</p> <p>$$\mathrm{I}=\frac{8}{5} \int_\limits1^{\sqrt{2}} \mathrm{k}^6-2 \mathrm{k}^4+\mathrm{k}^2 \mathrm{dk}$$</p> <p>$$\mathrm{I}=\frac{8}{5}\left[\frac{\mathrm{k}^7}{7}-\frac{2 \mathrm{k}^5}{5}+\frac{\mathrm{k}^3}{3}\right]_1^{\sqrt{2}}$$</p> <p>$$\mathrm{I}=\frac{8}{5}\left[\frac{8 \sqrt{2}}{7}-\frac{8 \sqrt{2}}{5}+\frac{2 \sqrt{2}}{3}-\frac{1}{7}+\frac{2}{5}-\frac{1}{3}\right]$$</p> <p>$$\begin{aligned} &\mathrm{I}=\frac{8}{5}\left[\frac{22 \sqrt{2}}{105}-\frac{8}{105}\right]\\ &\therefore 525 \cdot \mathrm{I}=176 \sqrt{2}-64 \end{aligned}$$</p>