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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?

Shaun
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Winten
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  • See here: http://math.stackexchange.com/questions/476693/using-residue-theorem-to-evaluate-int-0-pi-sin2n-theta-d-theta/476708#476708 – Ron Gordon Nov 21 '13 at 08:36
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    If $n$ is odd, substitute $(1-\cos^2)$ for $\sin^2$ then use $u = \cos$. If $n$ is even, keep applying double angle identities until you get an odd power of cosine. – Eric Auld Nov 21 '13 at 08:36
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    The right hand side reads $\frac{\Gamma(\tfrac{1}{2})\ \cdot\ \Gamma(\tfrac{1}{2}+\tfrac{n}{2})}{\Gamma(\tfrac{1}{2}+\tfrac{1}{2}+\tfrac{n}{2})}=\mathrm B(\tfrac{1}{2},\tfrac{1}{2}+\tfrac{n}{2})$ and I think the proof of this relation is a common one. – Nikolaj-K Nov 21 '13 at 09:05

2 Answers2

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Denote $$ I_k(x)=\int\sin^k xdx\qquad J_k=\int_0^\pi \sin^k xdx $$ Integration by parts gives $$ \begin{align} I_k(x) &=-\sin^{k-1}(x)\cos (x)-\int \cos (x)\cdot(k-1)\sin^{k-2}(x)\cos (x)dx \\ &=-\sin^{k-1}(x)\cos (x)+(k-1)\int \sin^{k-2}(x)(1-\sin^2(x))dx \\ &=-\sin^{k-1}x\cos x+(k-1)I_{k-2}(x)-(k-1)I_k(x) \\ \end{align} $$ So we derive the recurrence formula $$ I_k(x)=-\frac{\sin^{k-1}(x)\cos (x)}{k}+\frac{k-1}{k} I_{k-2}(x) $$ 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 $$ J_{2k}=\frac{(2k-1)(2k-3)\cdot\ldots\cdot 3\cdot 1}{2k(2k-2)\cdot\ldots\cdot 4\cdot 2}\cdot \pi\\ J_{2k+1}=\frac{2k(2k-2)\cdot\ldots\cdot 2}{(2k+1)(2k-1)\cdot\ldots\cdot 3}\cdot 2 $$ From here you can verify that $$ J_n=\sqrt\pi\frac{\Gamma(\frac{1+n}{2})}{\Gamma(1+\frac{n}{2})} $$

Norbert
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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.

Claude Leibovici
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