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?
 A: $\int \sin^5 x dx = \int(1-\cos^2x)^2\sin x dx =-\int (1-2\cos^2x + \cos^4 x) d\cos x$
Set $t=\cos x$ and integrate.
A: $$\int \sin^5 x dx = \int (\sin^2 x)^2 \cdot \sin x dx = \int (1-\cos^2 x)^2 \sin x dx$$
Substutute $v = \cos x$, meaning $dv = -\sin x dx$ and you get
$$\int (1 - \cos ^2 x)^2 x \cdot \sin x dx = -\int (1-v^2)^2 dv.$$ Should be easier from here on. 
A: Since
$$\begin{align}\sin^5x&=(1/2i)^5(e^{ix}-e^{-ix})^5
 \\&={(-1/4)^2(e^{i2x}-2+e^{-i2x})}^2 (1/2i)(e^{ix}-e^{-ix})
 \\&=(1/16)(e^{i4x}+e^{-i4x}-4e^{i2x}-4e^{-i2x}+6)(1/2i)(e^{ix}-e^{-ix})
 \\&=(1/32i)(e^{i5x}-e^{i5x}-5e^{i3x}+5e^{-i3x}+10e^{ix}-10e^{-ix})
 \\&=(1/16)\sin5x-(5/16)\sin3x+(5/8)\sin x,\end{align}$$
you'll have
$$∫\sin^5 x dx =(-1/80)\cos5x+(5/48)\cos3x-(5/8)\cos x+C.$$
A: 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 :-)
