Question

${\int }_{-3\pi }^{3\pi }{\sin }^2\theta .{\sin }^{2}2\theta d\theta$ is

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

Answer 1

The correct option is C $\frac{3\pi }{2}$

$I=\underset{-3\pi }{\overset{3\pi }{\int }}{\sin }^2\theta .{\sin }^{2}2\theta d\theta$
We know that if $f$ is an even function
$\Rightarrow \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$

$I=2\underset{0}{\overset{3\pi }{\int }}(\sin \theta .\sin 2\theta {)}^{2}d\theta$
$I=\frac{1}{2}\underset{0}{\overset{3\pi }{\int }}(2\sin \theta .\sin 2\theta {)}^{2}d\theta$
we know that $2\sin A.\sin B=\cos (A-B)-\cos (A+B)$

$I=\frac{1}{2}\underset{0}{\overset{3\pi }{\int }}(\cos \theta -\cos 3\theta {)}^{2}d\theta$
$I=\frac{1}{2}\underset{0}{\overset{3\pi }{\int }}({\cos }^2\theta -2\cos 3\theta \cos \theta +{\cos }^{2}3\theta )d\theta$
Using the formula$2\cos A.\cos B=\cos (A+B)+\cos (A-B)$, we get
$I=\frac{1}{2}\underset{0}{\overset{3\pi }{\int }}(\frac{1+\cos 2\theta }{2}-(\cos 4\theta +\cos 2\theta )+\frac{1+\cos 6\theta }{2})d\theta$
$I=\frac{1}{4}\underset{0}{\overset{3\pi }{\int }}(2+\cos 6\theta -2\cos 4\theta -\cos 2\theta )d\theta$
$I=\frac{1}{4}[\underset{0}{\overset{3\pi }{\int }}2d\theta +\underset{0}{\overset{3\pi }{\int }}\cos 6\theta d\theta -\underset{0}{\overset{3\pi }{\int }}2\cos 4\theta d\theta -\underset{0}{\overset{3\pi }{\int }}\cos 2\theta d\theta ]$
$I=\frac{1}{4}[2\times (3\pi -0)+(0-0)-2\times (0-0)-(0-0\left)\right]$
$I=\frac{6\pi }{4}=\frac{3\pi }{2}$