Question

The value of the definite integral ${\int }_0^{\pi /2}\sin x\sin 2x\sin 3xdx$ is equal to:

• A. $\frac{1}{3}$
• B. $-\frac{2}{3}$
• C. $-\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

Answer: (D)

$\frac{1}{6}$

${\int }_0^{\pi /2}\sin x\sin 2x\sin 3xdx$

$\Rightarrow \frac{1}{2}{\int }_0^{\pi /2}2\sin x\sin 2x\sin 3xdx$

$\Rightarrow \frac{1}{2}{\int }_0^{\pi /2}2\sin x\sin 3x\sin 2xdx$

We know that $2\sin A\sin B=\cos (A-B)-\cos (A+B)$

$\Rightarrow \frac{1}{2}{\int }_0^{\pi /2}\left(\cos \right(x-3x)-\cos (x+3x\left)\right)\sin 2xdx$

$\Rightarrow \frac{1}{2}{\int }_0^{\pi /2}(\cos 2x-\cos 4x)\sin 2xdx$

$\Rightarrow \frac{1}{2}{\int }_0^{\pi /2}\sin 2x\cos 2xdx-\frac{1}{2}{\int }_0^{\pi /2}\sin 2x\cos 4xdx$

$\Rightarrow \frac{1}{4}{\int }_0^{\pi /2}2\sin 2x\cos 2xdx-\frac{1}{4}{\int }_0^{\pi /2}2\sin 2x\cos 4xdx$

We know that $2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$\Rightarrow \frac{1}{4}{\int }_0^{\pi /2}\left(\sin \right(2x+2x)+\sin (2x-2x\left)\right)dx-\frac{1}{4}{\int }_0^{\pi /2}\left(\sin \right(2x+4x)+\sin (2x-4x\left)\right)dx$

$\Rightarrow \frac{1}{4}{\int }_0^{\pi /2}\sin 4xdx-\frac{1}{4}{\int }_0^{\pi /2}(\sin 6x-\sin 2x)dx$

$\Rightarrow \frac{1}{4}{\int }_0^{\pi /2}\sin 4xdx-\frac{1}{4}{\int }_0^{\pi /2}\sin 6xdx+\frac{1}{4}{\int }_0^{\pi /4}\sin 2xdx$

$\Rightarrow \frac{1}{4}[\frac{\sin 4x}{4}{]}_0^{\pi /4}-\frac{1}{4}[\frac{-\sin 6x}{6}{]}_0^{\pi /4}+\frac{1}{4}[\frac{-\cos 2x}{2}{]}_0^{\pi /4}$

$\Rightarrow \frac{-1}{12}+\frac{1}{4}=\frac{1}{6}$