Integrals — Class 12 Maths Solution

: Let $I = \int {{{\sin }^4}x\,dx} = {\int {\left( {\cfrac{{1 - \cos 2x}}{2}} \right)} ^2}dx$

$= \cfrac{1}{4}\int {\left( {1 + {{\cos }^2}2x - 2\cos x\,2x} \right)dx}$

$= \cfrac{1}{4}\int {\left[ {1 + \cfrac{{1 + \cos 4x}}{2} - 2\cos 2x} \right]} dx$

$= \cfrac{1}{4}\int {\left( 1 \right)dx} + \cfrac{1}{8}\int {\left( {1 + \cos 4x} \right)dx} - \cfrac{2}{4}\int {\cos 2x} dx$

$= \cfrac{3}{8}\int {\left( 1 \right)dx} + \cfrac{1}{8}\int {\cos 4x\,dx} - \cfrac{1}{2}\int {\cos \,2x\,dx}$

$= \cfrac{3}{8}x + \cfrac{1}{{32}}\sin 4x - \cfrac{1}{4}\sin 2x + C$