Question

The value of the integral ${\int }_{-3\pi }^{3\pi }\left|si{n}^{3}x\right|dx$ is equal to

• A. $\pi$
• B. $8\pi$
• C. $1$
• D. $8$

Answer 1

Answer: (D)

The period of the function $\left|{\sin }^{3}x\right|$ is $\pi ,$ thus $I=6{\int }_0^{\pi }{\sin }^{3}xdx$ $\Rightarrow I=6{\int }_0^{\pi }\frac{3\sin x-\sin (3x)}{4}dx($ As $\left(\sin (3x)=3\sin x-4{\sin }^{3}x\right))$
$\Rightarrow I=\frac{6}{4}{\int }_0^{\pi }(3\sin x-\sin (3x))dx$
$\Rightarrow I=\frac{3}{2}\left([-3\cos x{]}_0^{\pi }+{\left[\frac{\cos (3x)}{3}\right]}_0^{\pi }\right)$
$=\frac{3}{2}\left(3-(-3)+\left(\frac{-1}{3}-\frac{1}{3}\right)\right)$
$=\frac{3}{2}\left(6-\frac{2}{3}\right)=\frac{3}{2}\times \frac{16}{3}=8$