Is there any trig identity which can be used for $\int \sin^4x$? What is $\int \sin^4x$?
I tried to use the identity $\sin^2x = \frac{1-\cos 2x}{2}$, but I'm stuck.
 A: use that $$\sin(x)^4=\frac{1}{8} (-4 \cos (2 x)+\cos (4 x)+3)$$
A: Hint:
Square both sides
and use $$\cos2y=2\cos^2y-1$$
A: Hint:
Use that identity twice, so you get $$\frac14\int (1-\cos(2x))^2 \,dx$$Then expand the brackets and use another similar formula for the $\cos^2$ term.
A: Hint:
use
$$\sin^2ax=\left(\frac{1-\cos( 2ax)}{2}\right)$$
$$\cos^2ax=\left(\frac{1+\cos (2ax)}{2}\right)$$
A: Using the substitution above:
$\sin^4 x = (\frac {1-\cos 2x}2)^2 = \frac 14 - \frac 12 \cos 2x + \frac 14\cos^2 2x $ 
$\cos^2 2x = \frac {1+\cos 4x}{2}$ 
Or, using a little complex analysis
$\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\sin^4 x = \left(\frac {e^{ix} - e^{-ix}}{2i}\right)^4\\
\frac {e^{4ix} - 4e^{3ix}e^{-ix} + 6e^{2ix}e^{-2ix} - 4e^{ix}e^{-3ix} + 4e{-4ix})}{16}\\
\frac {(e^{4ix} + e^{-4ix})- 4(e^{2ix}+2e^{-2ix}) +6}{16}\\
\frac 18 \cos 4x - \frac 12 \cos 2x + \frac {3}{8}$ 
A: You can also calculate as follows:
$$\sin^4x=\sin^2x(1-\cos^2 x)=\sin^2x-\frac 14 \sin^2 2x=\frac {1-\cos 2x}2-\frac {1-\cos 4x}8$$where the second step uses $\sin 2x =2\sin x\cos x$ and you use your original identity for each of the $\sin^2$ expressions which result.
