Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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}

Posted by

infoexpert24

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 supplementary exams 2026 datesheet out; test on July 28
anam-khan

Centre, CBSE, NCERT defend NEP three-language policy in SC; seek dismissal of pleas against framework
anam-khan

Supreme Court to hear plea against CBSE evaluation policy for Gulf Class 12 students on July 14

Latest Question