Separate numerator and denominator integral I was given $$ \frac{\int_0^{\pi/2}\sin^{\sqrt{2}+1}(x) \, dx}{  \int_0^{\pi/2}\sin^{\sqrt{2}-1}(x) \, dx} $$
How to evaluate this integral?
Since denominator and numerator are different Integral substitution will fail , I tried with by part Integral but it lead to nowhere help?  
 A: Hint: For positive numbers $u,v$ one has
$$ \int_{0}^{\frac{\pi}{2}}(\sin x)^{2u-1}(\cos x)^{2v-1}dx=\frac{1}{2}B(u,v)$$
where $B(u,v)$ is Beta function, see http://en.wikipedia.org/wiki/Beta_function
Since in
$$\int_{0}^{\frac{\pi}{2}}(\sin x)^{\sqrt{2}+1}dx $$
one has $\sqrt{2}+1=2u-1$ and $0=2v-1$ we have
$$\int_{0}^{\frac{\pi}{2}}(\sin x)^{\sqrt{2}+1}dx=\frac{1}{2}B(\frac{\sqrt{2}}{2}+1,\frac{1}{2}).$$
Similarly,
$$\int_{0}^{\frac{\pi}{2}}(\sin x)^{\sqrt{2}-1}dx=\frac{1}{2}B(\frac{\sqrt{2}}{2},\frac{1}{2}).$$
Function Beta can be expressed by function Gamma (see http://en.wikipedia.org/wiki/Gamma_function) as follows: $B(u,v)=\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}$.
Hence
$$ \frac{\int_{0}^{\frac{\pi}{2}}(\sin x)^{\sqrt{2}+1}dx}{\int_{0}^{\frac{\pi}{2}}(\sin x)^{\sqrt{2}-1}dx}=\frac{\frac{1}{2}B(\frac{\sqrt{2}}{2}+1,\frac{1}{2})}{\frac{1}{2}B(\frac{\sqrt{2}}{2},\frac{1}{2})}=\frac{\Gamma(\frac{\sqrt{2}}{2}+1)\Gamma(\frac{1}{2})\Gamma(\frac{\sqrt{2}}{2}+\frac{1}{2})}{\Gamma(\frac{\sqrt{2}}{2}+\frac{3}{2})\Gamma(\frac{\sqrt{2}}{2})\Gamma(\frac{1}{2})}.$$
Now we use the following property of function Gamma: $\Gamma(t+1)=t\Gamma(t)$. We simplify the above result to
$$ \frac{\sqrt{2}}{\sqrt{2}+1}.$$
I hope that I didn't make any mistake in my calculations. 
