Definite Integrals — Maths STD 12 Science — Question

Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$

When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$

$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$

$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\sqrt{\text{t}}\text{ dt}$

$\Rightarrow\text{I}=\Bigg[\frac{2\text{t}^{\frac{3}{2}}}{3}\Bigg]^{\frac{\pi}{4}}_0$

$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{\pi}{4}\Big)^{\frac{3}{2}}$

$\Rightarrow\text{I}=\frac{1}{12}\pi^{\frac{3}{2}}$