A $\sin^n x$ integral
By trying to derive volume of N-sphere I came the integrals like: $\int_0^{\pi} \sin^n x dx $
Wolfram Mathematica was able integrate it giving the following: $$\int_0^{\pi} \sin^n x dx = \sqrt\pi\frac{\Gamma(\frac{1+n}{2})}{\Gamma(1+\frac{n}{2})} $$
But I have no clue of how this could be deduced, any ideas?
$\endgroup$ 32 Answers
$\begingroup$Denote$$ I_k=\int\sin^k xdx\qquad J_k=\int_0^\pi \sin^k xdx $$Integration by parts gives$$ \begin{align} I_k &=-\sin^{k-1}x\cos x-\int \cos x(k-1)\sin^{k-2}x\cos xdx\\ &=-\sin^{k-1}x\cos x+(k-1)\int \sin^{k-2}x(1-\sin^2 x) xdx\\ &=-\sin^{k-1}x\cos x+(k-1)I_{k-2}(x)-(k-1)I_k(x)\\ \end{align} $$So we derive the recurrence formula$$ I_k=-\frac{\sin^{k-1}\cos x}{k}+\frac{k-1}{k} I_{k-2}(x) $$Hence the desired integral is given by the recurrence formula$$ J_k=\frac{k-1}{k}J_{k-2} $$Note that $J_1=2$, $J_2=\pi/2$. From here you can easily verify that$$ J_{2k}=\frac{(2k-1)(2k-3)\cdot\ldots\cdot 3}{2k(2k-2)\cdot\ldots\cdot 4}\cdot 2\\ J_{2k+1}=\frac{2k(2k-2)\cdot\ldots\cdot 2}{(2k+1)(2k-1)\cdot\ldots\cdot 3}\cdot \frac{\pi}{2} $$From here you can verify that$$ J_n=\sqrt\pi\frac{\Gamma(\frac{1+n}{2})}{\Gamma(1+\frac{n}{2})} $$
$\endgroup$ $\begingroup$If you use the tangent half-angle substitution, the antiderivative involves the hypergeometric 2F1 function. Taking the values at the limits gives what you report.
If I properly remember, Sin[x]^n can be represented as a linear combination of sigle powers of Cos (if n is even) and as a linear combination of sigle powers of Sin (if n is odd). This probably could help.
