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)

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 Class 12 result verification 2024 application out at cbse.gov.in; last date, fees
anam-khan

CBSE results 2024 later for 102 students over fake certificates; board issues show cause notice
anam-khan

What after CBSE Class 12 results 2024? From CUET UG to JEE Advanced, all you need to know

Latest Question