Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Evaluate the integrals in Exercises 1 to 8 using substitution. sin^{-1}2x/ (1+x^2)

Evaluate the integrals in Exercises 1 to 8 using substitution.

    Q3.    \int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx

Answers (1)

best_answer

\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx
let x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta
when x = 0 then \theta= 0 and when x = 1 then \theta= \pi/4

\\=\int_{0}^{\pi/4}\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}2\theta \sec^2\theta d\theta\\
Taking \theta as a first function and \sec^2\theta as a second function, by using by parts method

\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]^{\pi/4}_0\\ =2[\pi/4+\log(1/\sqrt{2})]\\ =\pi/4-\log 2

Posted by

manish

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 practical exams 2026 will be cancelled over non-compliance of guidelines; board issues SOPs

Latest Question