Question
By using the properties of definite integral, evaluate the integral in Exercise:
$ \int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\sin^{2}\text{x}\ \text{dx}$
$ \int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\sin^{2}\text{x}\ \text{dx}$
$\Rightarrow\ \ \text{I}=2\int\limits_{0}^{\frac{\pi}{2}}\sin^{2}\bigg(\frac{\pi}{2}-\text{x}\bigg)\text{dx}\ \ \ \Big[\because\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x)}\text{dx}=\Big]$
$\Rightarrow\ \ \text{I}=2\int^{\frac{\pi}{2}}\limits_{0}\cos^{2}\text{x}\ \text{dx}$
Adding eq. (i) and (ii),
$21=2\int^{\frac{\pi}{2}}\limits_{0}\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)\text{dx}=2\int^{\frac{\pi}{2}}\limits_{0}1\text{dx}=2\text{(x)}^{\frac{\pi}{2}}_{0}=2.\frac{\pi}{2}=\pi$
$\Rightarrow\ \ \ \text{I}=\frac{\pi}{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.