How to evaluate the following integral using hypergeometric function?

Question

May I know how this integral was evaluated using hypergeometric function? $$\int \sin^n x\ dx$$

Wolframalpha showed this result but with no steps

Thanks in advance.

Answer 1

Assuming $n$ is a non-negative integer, you could use binomial theorem: $$\begin{eqnarray} \sin^n(x) &=& \left( \frac{\exp(i x) - \exp(-i x)}{2i}\right)^n = \frac{1}{2^n i^n} \sum_{m=0}^n \binom{n}{m} (-1)^m \exp\left( i \left(n-2m\right)x \right) \\ &=& \frac{1}{2^n i^n} \sum_{m=0}^n \binom{n}{m} (-1)^m \left(\cos\left(\left(n-2m\right)x \right) + i \sin\left( \left(n-2m\right)x \right) \right) \end{eqnarray} $$ Since the left-hand-side is real we only keep cosines for even $n$: $$\begin{eqnarray} \sin^{2n}(x) &=& \frac{1}{2^{2n}} \sum_{m=0}^{2n} \binom{2n}{m}\left(-1\right)^{n-m} \cos\left(\left(2n-2m\right)x\right) \\ &\stackrel{\text{symmetry}}{=}& \frac{1}{2^{2n}} \binom{2n}{n} + \frac{1}{2^{2n-1}} \sum_{m=0}^{n-1} \binom{2n}{m}\left(-1\right)^{n-m} \cos\left(2 \left(n-m\right)x\right) \\ &\stackrel{m\to n-m} =& \frac{1}{2^{2n}} \binom{2n}{n} + \frac{1}{2^{2n-1}} \sum_{m=1}^n \binom{2n}{n+m} (-1)^n \cos(2 m x) \tag{1} \end{eqnarray}$$ and, likewise, only sines for odd $n$: $$ \sin^{2n+1}(x) = \frac{1}{2^{2n}} \sum_{m=0}^n \binom{2n+1}{n+1+m} (-1)^m \sin\left((2m+1)x\right) \tag{2} $$ We can now integrate element-wise: $$ \int \sin^{2n}(x) \, \mathrm{d}x = \frac{1}{2^{2n}} \binom{2n}{n} x + \frac{1}{2^{2n-1}} \sum_{m=1}^n \binom{2n}{n+m} (-1)^n \frac{\sin(2 m x)}{2m} + \text{const.} $$ $$ \int \sin^{2n+1}(x) \, \mathrm{d}x = -\frac{1}{2^{2n}} \sum_{m=0}^n \binom{2n+1}{n+1+m} (-1)^m \frac{\cos\left((2m+1)x\right)}{2m+1} + \text{const.} $$

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To obtain a hypergeometric function, let $u = \sin(x)$. Then $$ \int \sin^n(x)\, \mathrm{d}x = \int \frac{u^n}{\sqrt{1-u^2}} \mathrm{d}u $$ Now see this answer of mine on how to find the anti-derivative of $\int u^a (1-u)^b \mathrm{d} u$. Applying the same principles, we find: $$ \int \frac{u^n}{\sqrt{1-u^2}} \mathrm{d}u =\int u^n \cdot {}_1F_0\left(\left.\begin{array}{c} \frac{1}{2} \\ - \end{array} \right| u^2 \right) \mathrm{d} u = \int \frac{\mathrm{d}}{\mathrm{d}u} \left( \frac{u^{n+1}}{n+1} \cdot {}_2F_1\left(\left.\begin{array}{cc} \frac{1}{2} & \frac{n+1}{2} \\ & \frac{n+3}{2} \end{array} \right| u^2 \right) \right) \mathrm{d} u $$ Thus, we have: $$ \int \sin^n(x) \, \mathrm{d}x = \frac{\sin^{n+1}(x)}{n+1} \cdot {}_2F_1\left(\left.\begin{array}{cc} \frac{1}{2} & \frac{n+1}{2} \\ & \frac{n+3}{2} \end{array} \right| \sin^2(x) \right) + \text{const.} \tag{3} $$ This works where $u = \sin(x)$ is invertible. To extend validity of the answer, differentiate it. We would get $$ \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\sin^{n+1}(x)}{n+1} \cdot {}_2F_1\left(\left.\begin{array}{cc} \frac{1}{2} & \frac{n+1}{2} \\ & \frac{n+3}{2} \end{array} \right| \sin^2(x) \right) \right) = \frac{\sqrt{\cos^2(x)}}{\cos(x)} \sin^{n} (x) $$ and since the pre-factor $\frac{\sqrt{\cos^2(x)}}{\cos(x)}$ is a differential constant, i.e. its derivative is zero, we arrive at: $$ \int \sin^n(x) \, \mathrm{d}x = \frac{\cos(x)}{\sqrt{\cos^2(x)}} \frac{\sin^{n+1}(x)}{n+1} \cdot {}_2F_1\left(\left.\begin{array}{cc} \frac{1}{2} & \frac{n+1}{2} \\ & \frac{n+3}{2} \end{array} \right| \sin^2(x) \right) + \text{const.} \tag{4} $$ This can be related to the answer provided by Wolfram|Alpha, and thus by Mathematica, using Kummer's relations.