I tried different methods of integration, but I can not solve this problem.
$$\int \sin^n (x) dx$$
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3Have you heard of reduction formulae? – Edward Evans Oct 14 '16 at 19:45
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Can we assume that $n \in \mathbb N$? – Doug M Oct 14 '16 at 20:08
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Hm, sadly, the link only explains how to do it for a given $n$ and cannot be easily made to work for any $n$. – Simply Beautiful Art Oct 14 '16 at 20:24
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Hint: $${\displaystyle \sin ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}(-1)^{({\frac {n-1}{2}}-k)}{\binom {n}{k}}\sin {{\big (}(n-2k)\theta {\big )}}}\quad {\text{where n is odd}}$$ $${\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{({\frac {n}{2}}-k)}{\binom {n}{k}}\cos {{\big (}(n-2k{\big )}\theta )}}\quad {\text{where n is even}}$$
E.H.E
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3One might reasonable request a source for this identity, which can be easily obtained applying the binomial theorem to $$\sin^n(\theta)=\left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^n$$ – Mark Viola Oct 14 '16 at 19:53
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When $n$ is even, how should one interpret the term $(n-1)/2$ in your expression? – Mark Viola Oct 14 '16 at 20:10
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@Dr.MV you can use the other identity ${\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{({\frac {n}{2}}-k)}{\binom {n}{k}}\cos {{\big (}(n-2k{\big )}\theta )}}$ – E.H.E Oct 14 '16 at 20:15