Definite integration involving Trigonometry. $$ \frac{\int_0^{\pi/2} (\sin x)^{2½+1} \, dx}{\int_0^{\pi/2} (\sin x)^{2½-1} \, dx}$$
I'm puzzled. 
How to proceed...

 A: We have
$$\sin^{\sqrt{2}+1}x=\sin^{\sqrt{2}-1}x\sin^2x=\sin^{\sqrt{2}-1}x-\sin^{\sqrt{2}-1}x\cos^2x.$$
Now use integration by parts
$$\int f'g=fg-\int fg'$$
with 
$$f(x)=\frac{1}{\sqrt{2}}\sin^{\sqrt{2}}x,\quad g(x)=\cos x,$$
it follows that
$$\begin{aligned}&\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}x\cos^2x\\
=&\frac{1}{\sqrt{2}}\sin^{\sqrt{2}}x\cos x|_{x=0}^{\frac{\pi}{2}}+\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}}x\sin xdx\\
=&\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx
\end{aligned}$$
Therefore
$$\begin{aligned}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx&=\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}xdx-\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}x\cos^2xdx\\
&=\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}xdx-\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx,
\end{aligned}$$
which implies
$$\frac{\sqrt{2}+1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx=\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}xdx.$$
Hence
$$\frac{\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}+1}xdx}{\int_0^{\frac{\pi}{2}}\sin^{\sqrt{2}-1}xdx}=\frac{\sqrt{2}}{\sqrt{2}+1}=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}.$$
