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 12.

Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 12.

Answers (1)

\frac{\pi}{4}

Hint: We use indefinite integral formula then put limits to solve this integral.

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

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

Put \sin x=t

\cos x \; \; dx=dt

dx=\frac{dt}{\cos x}

When x=0  then t=0  and when x=\frac{\pi}{2}  then t=1

=\int_{0}^{1} \frac{1}{1+t^{2}} d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]

=[\tan ^{-1}(t)]^{1}_0

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

=\left [\frac{\pi}{4}-0 \right ]

= \frac{\pi}{4}

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 OSM row: Congress calls govt action “eyewash”, demands Pradhan’s removal
anam-khan

CBSE upgrades re-evaluation portal, extends session limits for students
anam-khan

CBSE gets new chairman, secretary; no central post for former secretary till December 2030

Latest Question