Integral $ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $ I have this difficult integral to solve.
$$ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $$
Now my approach is this: split $(\sin x)^{\sqrt 2 + 1}$ and $(\sin x)^{\sqrt 2 - 1}$ as $(\sin x)^{\sqrt 2}.(\sin x)$ and $(\sin x)^{\sqrt 2 - 2}.(\sin x)$ respectively, and then apply parts. But that doesn't seem to lead anywhere. Hints please!
Edit:
This is what I did (showing just for the numerator)
$$ \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx $$
$$ = \int_0^{\pi/2} (\sin x)^{\sqrt 2}.(\sin x) dx  $$
$$ = (-\cos x)(\sin x)^{\sqrt 2}\Bigg|_0^{\pi/2} + \int_0^{\pi/2}(\sin x)^{ \sqrt 2 - 1 }(\cos^2 x) dx  $$
(taking $ v = \sin x $ and $ u = (\sin x)^{\sqrt 2} $ in the $ \int uv $ formula)
$$ = \int_0^{\pi/2}\left( (\sin x)^{ \sqrt 2 - 1 } - (\sin x)^{ \sqrt 2 + 1 } \right) dx  $$
Similarly for the denominator. This does give a reduction formula but then I don't see how to really use it for finding the answer.
 A: Just like what @O.L pointed out, you can use Beta function to express the solution. In fact, you can change variables to get 
\begin{eqnarray*}
&&\frac{\int_0^{\pi/2}\sin^{\sqrt{2}+1}xdx}{\int_0^{\pi/2}\sin^{\sqrt{2}-1}xdx}=\frac{\int_0^{1}u^{\sqrt{2}+1}(1-u^2)^{-1/2}du}{\int_0^{1}u^{\sqrt{2}-1}(1-u^2)^{-1/2}du}\\
&=&\frac{\int_0^{1}t^{\frac{1}{2}(\sqrt{2}+1)}(1-t)^{-1/2}\frac{1}{2t^{1/2}}dt}{\int_0^{1}t^{\frac{1}{2}(\sqrt{2}-1)}(1-t)^{-1/2}\frac{1}{2t^{1/2}}dt}=\frac{\int_0^{1}t^{\frac{1}{2}\sqrt{2}}(1-t)^{-1/2}dt}{\int_0^{1}t^{\frac{1}{2}(\sqrt{2}-2)}(1-t)^{-1/2}dt}\\
&=&\frac{B(\frac{\sqrt{2}+2}{2},\frac{1}{2})}{B(\frac{\sqrt{2}}{2},\frac{1}{2})}=\frac{\Gamma(\frac{\sqrt{2}+2}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{\sqrt{2}+3}{2})}\frac{\Gamma(\frac{\sqrt{2}+1}{2})}{\Gamma(\frac{\sqrt{2}}{2})\Gamma(\frac{1}{2})}=\frac{\Gamma(\frac{\sqrt{2}+2}{2})\Gamma(\frac{\sqrt{2}+1}{2})}{\Gamma(\frac{\sqrt{2}+3}{2})\Gamma(\frac{\sqrt{2}}{2})}\\
&=&\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}+1}{2}}=2-\sqrt{2}.
\end{eqnarray*}
A: Let the top integral be $I$, and the bottom one $J$. To make typing simpler, let $a=\sqrt{2}$.
Integrate the top one by parts, letting $du=\sin x$ and $v=\sin^{a} x$.  This is the standard way to get a reduction formula for $\int \sin^n x\,dx$.
So $dv=a\cos x\sin^{a-1}x\,dx$ and we can take $u=-\cos x$.  Then
$$I=\left. -\cos x \sin^{a}x\right|_0^{\pi/2}+\int_0^{\pi/2}a\cos^2 x\sin^{a-1} x\,dx.$$
The first part dies at both ends. Rewrite $\cos^2 x$ as $1-\sin^2 x$. Then 
$$I=aJ -aI.$$
Now we get 
$$I=\frac{a}{a+1}J$$
and it's over. 
A: Hint:
$$\int_0^{\pi/2}\sin^{2p-1}x\,\cos^{2q-1} x\,dx=\frac{1}{2}B(p,q)=\frac{\Gamma(p)\Gamma(q)}{2\Gamma(p+q)}.$$
Using this formula, we obtain
$$\frac{\int_0^{\pi/2}\sin^{\sqrt{2}+1}x\,dx}{\int_0^{\pi/2}\sin^{\sqrt{2}-1}x\,dx}=\frac{B(1+\frac{\sqrt{2}}{2},\frac12)}{B(\frac{\sqrt{2}}{2},\frac12)}=\frac{\Gamma(1+\frac{\sqrt{2}}{2})\Gamma(\frac{1+\sqrt{2}}{2})}{\Gamma(\frac{\sqrt{2}}{2})\Gamma(\frac{3+\sqrt{2}}{2})}=\frac{\sqrt{2}}{1+\sqrt{2}}=2-\sqrt{2}.$$
