Question

Evaluate the integral
${\int }_0^{2\pi }x.{\sin }^{2}xdx$

• A. $\pi$
• B. $\pi /2$
• C. ${\pi }^2$
• D. 0

Answer 1

The correct option is C ${\pi }^2$

Given, ${\int }_0^{2\pi }x.{\sin }^{2}xdx=I\left(say\right)$
We know that ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(\right(a+b)-x)dx$
Hence , ${\int }_0^{2\pi }(2\pi -x){\sin }^2(2\pi -x)dx=I$
So, ${\int }_0^{2\pi }\left(2\pi \right){\sin }^{2}xdx=2I$
$\Rightarrow$ ${\int }_0^{2\pi }\left(\pi \right){\sin }^{2}xdx=I$
$\Rightarrow$ ${\int }_0^{2\pi }\left(\pi \right)\frac{(1-\cos 2x)}{2}dx=I$
$\Rightarrow$ $I=\pi \left(\right[\frac{2\pi }{2}-\frac{\sin 2\left(2\pi \right)}{4}]-[0\left]\right)={\pi }^2$
Hence, option 'C' is correct.