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 19 Definite Integrals Exercise Revision Exercise Question 5 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 5 Maths Textbook Solution.

Answers (1)

best_answer

Answer:  \frac{\pi}{4}-\frac{1}{2} \log 2

Given:  \int_{0}^{1} \tan ^{-1} x d x

Hint: Use the integration by parts method

Solution:

\int_{0}^{1} \tan ^{-1} x d x

\int_{0}^{1} \tan ^{-1} x \cdot 1 d x

We know that  \int u v d x=u \int v d x-v \int u d x

\begin{aligned} &=\left(x\tan ^{-1} x \right)_{0}^{1}-\int_{0}^{1} \frac{1}{1+x^{2}} \times x d x \\ & \end{aligned}

=\frac{\pi}{4}-\frac{1}{2} \int_{0}^{1} \frac{1}{t} d t \\

=\frac{\pi}{4}-\frac{1}{2}[\log (t)]_{0}^{1}=\frac{\pi}{4}-\frac{1}{2}\left[\log \left(1+x^{2}\right)\right]_{0}^{1}

\begin{aligned} &=\frac{\pi}{4}-\frac{1}{2}(\log 2-\log 1) \\ & \end{aligned}                                       (\because log 1= 0)

=\frac{\pi}{4}-\frac{1}{2} \log 2

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

Latest Question