Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 13

Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 13

Answers (1)

best_answer

Answer: \frac{\pi}{4}

Given:  \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1+\sin ^{2} x} d x

Hint: You must know about the integration of  \frac{1}{1+x^{2}}

Solution:  \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1+\sin ^{2} x} d x

Let \sin x=t

\cos xdx=dt    (Differentiate w.r.t to x)

Now  \int_{0}^{\frac{\pi}{2}} \frac{1}{1+t^{2}} d t

\begin{aligned} &=\left(\tan ^{-1} t\right)_{0}^{\frac{\pi}{2}}=\left[\tan ^{-1}(\sin x)\right]_{0}^{\frac{\pi}{2}} \\ & \end{aligned}

=\tan ^{-1} 1-\tan ^{-1} 0 \\

=\frac{\pi}{4}-0\left(\because \tan \frac{\pi}{4}=1\right)

Posted by

infoexpert27

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 extends single girl child scholarship registration deadline till November 20
anam-khan

CBSE reminds schools to submit registration data of Class 9, 11 students by October 16
anam-khan

CBSE reopens online portal for submission of LOC for Class 10, 12 board exams 2026; submit forms by Oct 8

Latest Question