Integrate $\sin^5 x$ I know $\sin x$ integrates to $-\cos x$ but ive never seen $\sin^5(x)$ integrated. 
would I need to expand it to $\sin x\sin x\sin x\sin x\sin x$ and then how would I complete the integration from here?

Here's another answer:
We know from using integration by parts that
$$
\int\sin^5x dx = -\frac{1}{5}\sin^4x\cos x + \frac{4}{5}\int\sin^3x dx
$$
Applying intgration by parts to $\int\sin^3xdx$ means
$$
\int\sin^5x dx = - \frac{1}{5}\sin^4x\cos x + \frac{4}{5}(-\frac{3}{4}\cos x + \frac{1}{12}\cos 3x)
$$
$$
= - \frac{1}{5}\sin^4x\cos x - \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}(\sin^2x)^2\cos x- \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}\cos x(1 - \cos^2 x)^2- \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}\cos x(1 - 2\cos^2 x + \cos^4 x) - \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}\cos x + \frac{2}{5}\cos^3 x - \frac{1}{5}\cos^5 x- \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}\cos x + \frac{2}{5}(\frac{3}{4}\cos x + \frac{1}{4}\cos 3x) - \frac{1}{5}(\frac{5}{8}\cos x + \frac{5}{16}\cos 3x + \frac{1}{16}\cos 5x) - \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= \cos x(-\frac{1}{40} - \frac{3}{5}) + \cos 3x(\frac{1}{15} + \frac{3}{80}) - \frac{1}{80}\cos 5x
$$
$$
= -\frac{5}{8}\cos x + \frac{5}{48}\cos 3x - \frac{1}{80}\cos 5x
$$
However, given I obtained this problem from Apostol "Calculus" Volume 1 Section 5.10 Question 10 (c) (Page 221) and the problem in Apostol states
$$
\int\sin^5x dx = -\frac{5}{8}x + \frac{5}{48}\cos 3x - \frac{1}{8}\cos 5x
$$
I can only assume that I found a typographical error in Apostol's epic tome! Hope this helps :-)
