I need to find an expression for $\int\sin^n(x)dx$ with $n\in\mathbb{N}$. I am interested in answers that tell me how to start solving this problem, as much as I am interested in the actual solution.
For now I have found the following identity: \begin{equation} \int\sin^n(x)=\dfrac{n-1}{n}\int\sin^{n-2}(x)dx-\dfrac{cos(x)sin^{n-1}(x)}{n} \end{equation} which I used to find a formula for every even $n$ (I have not verified this equation, so correct me if I am wrong):
For every $n=2k, k\in\mathbb{N}$: \begin{equation} \int\sin^n(x)dx=x\cdot\prod_{i=0}^{k}{\left(\dfrac{1}{n-2i}\right)}\prod_{i=i}^{k}{\left(n-(2i-1)\right)}-\sum_{i=1}^{2k}\dfrac{cos(x)sin^{(n-(2i-1))}(x)}{n-(2i-2)} \end{equation} The formula for uneven natural numbers $n=2k+1$ would look the same, just substituting the first $x$ with $-\cos(x)$. However I don't see of to use these results for a general formula.
Considering the complexity of this expression and the fact that I have to solve this for school, I suspect that I am missing an easy(/easier) way of solving this and a more compact solution. I would be very grateful for any answer giving me an idea of how I may solve this problem (, or the solution).
EDIT - Thanks to user26977's answer I got an answer for odd exponents. (Please, tell me if you see any error!) \begin{eqnarray} \int\sin^{2n+1}(x)dx=\int\sin(x)sin^{2n}(x)dx \end{eqnarray} Now, using the substitution rule with $t=cos(x)$ we get \begin{eqnarray} \int\sin(x)sin^{2n}(x)dx &=& -\int (1-t^2)^n dt = -\int\left(\sum_{k=0}^n\binom{n}{k}(-t^2)^k\right)dt\\ &=&-\sum_{k=0}^n\left(\binom{n}{k}(-1)^k\int t^{2k}dt\right) = -\sum_{k=0}^n\left(\binom{n}{k}(-1)^k\dfrac{t^{2k+1}}{2k+1}\right)\\ &=& \sum_{k=0}^n\left(\binom{n}{k}(-1)^k\dfrac{\cos^{2k+1}(x)}{2k+1}\right) \end{eqnarray} Would this be correct? Isn't there any more compact form?
EDIT 2 - Just for the record, the formula I gave above for even $n$ is incomplete.