By writing $\displaystyle \sin^{2n} (x) = \sin^{2n-2}( x ) [1- \cos^2( x )] $, use integration by parts (just once, in the second term) to derive the reccurence relation:

$\displaystyle I_{2n} = \left(\frac{2n-1}{2n}\right)I_{2n-2}$ $\displaystyle (n \ge 1)$, for $\displaystyle I_{2n} = \int_{0}^{\pi} \sin^{2n}( x )dx$.

Hence find the value of $\displaystyle I_4$ without using your calculator at any stage.