Integrals — Class 12 Maths Solution

Let $I = \int\limits_{ - \pi /2}^{\pi /2} {{{\sin }^2}x} dx$

Let $f\left( x \right) = {\sin ^2}x$
$\Rightarrow$ $f\left( x \right) = f\left( { - x} \right)$

$\therefore$ $f\left( x \right)$ is an even function.
$\therefore$ $I = 2\int\limits_0^{\pi /2} {{{\sin }^2}xdx}$

$= 2\int\limits_0^{\pi /2} {\left( {\cfrac{{1 - \cos 2x}}{2}} \right)} dx = \left[ {x - \cfrac{{\sin 2x}}{2}} \right]_0^{\pi /2} = \left[ {\cfrac{\pi }{2} - \cfrac{{\sin \pi }}{2}} \right]$

$\Rightarrow$ $I = \cfrac{\pi }{2}$