Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Integrate the functions sin ^ -1 2 x / (1+x^2)

Q22  Integrate the functions \sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )

Answers (1)

best_answer

\sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )

\int \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx
let x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta
 

\\=\int\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int2\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]\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]+C\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]+C\\ =2x\tan^{-1}x+2\log (1+x^2)^{-1/2}\\ =2x\tan^{-1}x-\log(1+x^2)+C

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 with guidelines, says board

Latest Question