Question-97892

Question Number 97892 by pranesh last updated on 10/Jun/20

Answered by john santu last updated on 10/Jun/20

\sin^{2} x = \frac{1}{2} - \frac{1}{2} \cos 2 x

\sin^{4} x = \frac{1}{4} - \frac{1}{2} \cos 2 x + \frac{1}{8} + \frac{1}{8} \cos 4 x

= \frac{3}{8} - \frac{1}{2} \cos 2 x + \frac{1}{8} \cos 4 x

\int \sin^{4} x dx = \frac{3}{8} x - \frac{1}{4} \sin 2 x + \frac{1}{32} \sin 4 x + c