Solving the integral of $sin^n(x)$

Question

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.

Answer 1

Hint:

If you don't like the recurrence relation, you can address this integral using the complex representation.

Let $$\sin(x)=\frac{z-z^{-1}}{2i}$$ where $z=e^{i\theta}$, and $dz=ie^{i\theta}d\theta=izd\theta$.

The integral becomes

$$\int\left(\frac{z-z^{-1}}{2i}\right)^n\frac{dz}{iz},$$

and by the Binomial development,

$$\frac1{i(2i)^n}\int\sum_{k=0}^n(-1)^{n-k}\binom nkz^{2k-n-1}dz.$$

After integration,

$$\frac1{i(2i)^n}\sum_{k=0}^n(-1)^{n-k}\binom nk\frac{z^{2k-n}}{2k-n},$$ which is

$$\frac1{i(2i)^n}\sum_{k=0}^n(-1)^{n-k}\binom nk\frac{\cos((2k-n)\theta+i\sin((2k-n)\theta)}{2k-n}.$$

Depending on the parity of $n$, the real of imaginary terms will cancel each other in pairs to yield a real result.

Also note that for even $n$, the central term has the exponent $z^{-1}$, which exceptionally integrates as $\ln(z)$, i.e. $i\theta$.

Answer 2

For odd $n$ you can write the integral as

$$\int \sin^{2k+1} x\; \mathrm dx = \int \sin^{2k} x \sin x\;\mathrm dx = - \int (1-t^2)^{k} \;\mathrm dt$$ and then use change of variable $t = \cos x$

Answer 3

For even $n=2k\geq2$ use integration by part to write $$ I=\int\sin^{2k}x\,dx=-\sin^{2k-1}x\cos x+(2k-1)\int\sin^{2k-2}\cos^2x\,dx $$ and the basic trig relation $\sin^2x+\cos^2x=1$ allows to write the last expression as $$ I=\sin^{2k-1}x\cos x-(2k-1)\int\sin^{2k}x\,dx+(2k-1)\int\sin^{2k-2}x\,dx $$ i.e. $$ I=\frac1{2k}\left[\sin^{2k-1}x\cos x+(2k-1)\int\sin^{2k-2}x\,dx\right] $$ Repeat the procedure $k-1$ more times.