Question

Find the integrals of the function:

sin⁴ x

Answer 1

Let ${\displaystyle I=\int {\sin }^{4}x dx}$

${\displaystyle =\int {\left(\frac{1-\cos 2x}{2}\right)}^2 dx}$

${\displaystyle =\frac{1}{4}\int (1+{\cos }^{2}2x-2\cos 2x)dx}$

${\displaystyle =\frac{1}{4}\int [1+\frac{1+\cos 4x}{2}-2\cos 2x]dx}$

${\displaystyle =\frac{1}{4}\int 1dx+\frac{1}{8}\int (1+\cos 4x) dx-\frac{2}{4}\int \cos 2x dx}$

${\displaystyle =\frac{3}{8}\int 1 dx+\frac{1}{8}\int \cos 4x dx-\frac{1}{2}\int \cos 2x dx}$

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