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

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

Answers (1)

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}

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