#### Provide solution for RD Sharma Maths Class12 Chapter Definite Integrals exercise 19.2 question 15.

Answer: $\frac{\pi \sqrt{\pi}}{12}$

Hint: We use indefinite formula then put limits to solve this integral.

Given: $\int_{0}^{1}\frac{\sqrt{\tan ^{-1}x} }{1+x^2}dx$

Solution: $I=\int_{0}^{1}\frac{\sqrt{\tan ^{-1}x} }{1+x^2}dx$

Put $\tan^{-1}x=t$

$\frac{1}{1+x^2}dx=dt$

$dx=(1+x^2)dt$

When x=0  then t=0  and

when x=1  then $t =\frac{\pi}{4}$

\begin{aligned} &I=\int_{0}^{\frac{\pi}{4}} \frac{\sqrt{t}}{1+x^{2}}\left(1+x^{2}\right) d t \\ &=\int_{0}^{\frac{\pi}{4}} \sqrt{t} d t \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} t^{\frac{1}{2}} d t \\ &=\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{\frac{\pi}{4}} \\ &=\left[\frac{t^{\frac{3}{2}}}{3}\right]_{0}^{\frac{\pi}{4}} \\ &=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{\frac{\pi}{4}} \\ &=\frac{2}{3}\left[\left(\frac{\pi}{4}\right)^{\frac{3}{2}}-0\right] \end{aligned}

\begin{aligned} &=\frac{2}{3}\left[\frac{\pi^{\frac{3}{2}}}{2^{2 \times \frac{3}{2}}}\right] \\ &=\frac{2}{3}\left[\frac{\pi \sqrt{\pi}}{2^{3}}\right] \\ &=\frac{2 \pi \sqrt{\pi}}{3 \times 8} \\ &=\frac{\pi \sqrt{\pi}}{12} \end{aligned}