$\displaystyle \int_0^\pi \! 29sin^4(3t) \, dt.$
Not sure if I have to substitute just once or twice. Any suggestions?
$\displaystyle \sin^4{u} =$
$\displaystyle (\sin^2{u})^2 =$
$\displaystyle \left[\frac{1 - \cos(2u)}{2}\right]^2 =$
$\displaystyle \frac{1}{4}\left[1 - 2\cos(2u) + \cos^2(2u)\right] =$
$\displaystyle \frac{1}{4}\left[1 - 2\cos(2u) + \frac{1 + \cos(4u)}{2}\right] =$
$\displaystyle \frac{3}{8} - \frac{\cos(2u)}{2} + \frac{\cos(4u)}{8}$
Before attempting integrals like the one you posted you should ensure that you are thoroughly familiar with all required trigonometric identities. Otherwise you have no chance.
For example, you should know that $\displaystyle \cos (2u) = \cos^2 u - \sin^2 u = 1 - 2 \sin^2 u$ ....