Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 6 maths textbook solution.

Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 6 maths textbook solution.

Answers (1)

\tan ^{-1}e-\frac{\pi}{4}

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

Given: \int_{0}^{1}\frac{e^x}{1+e^{2x}}dx

Solution: \int_{0}^{1}\frac{e^x}{1+e^{2x}}dx

Put e^x=t

e^x dx=dt

When x=0  then t=1  and when x=1  then t=e

\begin{aligned} &I=\int_{1}^{e} \frac{1}{1+t^{2}} d t \\ &=\left[\tan ^{-1} t\right]_{1}^{e} \\ &=\tan ^{-1} e-\tan ^{-1} 1\; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} 1=\frac{\pi}{4}\right] \end{aligned}

=\tan ^{-1}e-\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 practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question