Question

The value of ${\int }_0^{\pi }{\sin }^{4}xdx$ is:

• A. $\frac{3\pi }{16}$
• B. $\frac{3}{16}$
• C. $0$
• D. $\frac{3\pi }{8}$

Answer 1

The correct option is D $\frac{3\pi }{8}$

${\sin }^4\left(x\right)={\sin }^{2}x.{\sin }^{2}x={\sin }^{2}x(1-{\cos }^{2}x)$

$={\sin }^{2}x-{\sin }^{2}x{\cos }^{2}x$

$=\frac{1-\cos 2x}{2}-\frac{4{\sin }^{2}x.{\cos }^{2}x}{4}$

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

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

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

$=\frac{6}{16}+\frac{\cos 4x}{8}-\frac{\cos 2x}{2}$

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

Therefore

${\int }_0^{\pi }{\sin }^{4}xdx={\int }_0^{\pi }(\frac{3}{8}+\frac{\cos 4x}{8}-\frac{\cos 2x}{2})dx$

$={[\frac{3x}{8}+\frac{\sin 4x}{32}-\frac{\sin 2x}{4}]}_0^{\pi }$

$=\frac{3\pi }{8}$.