Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Choose the correct answers in Exercises 41 to 44. The value of is

Choose the correct answers in Exercises 41 to 44.

    Q44.    The value of \int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx is

                (A)    1

                (B)    0

                (C)    -1

                (D)    \frac{\pi}{4}

Answers (1)

best_answer

\\Let\ I=\int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx\\

\\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )\\ =tan^{-1}\left ( \frac{x-(1-x)}{1+x(1-x)} \right )\\ =tan^{-1}x-tan^{-1}(1-x)               as      tan^{-1}\left ( \frac{a-b}{1+ab} \right )=tan^{-1}a-tan^{-1}b

Now the integral can be written as

\\I=\int_{0}^{1} \left ( tan^{-1}x-tan^{-1}(1-x) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}(1-(1-x)) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}x\right )dx\\ I=-I\\ 2I=0\\ I=0

(B) is correct.

Posted by

Sayak

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 issues final reminder to schools for supplementary exam LOC submission; submit by today without late fee
anam-khan

CBSE Supplementary Exam 2025: Registrations open for Class 10, 12 private students today
anam-khan

CBSE opens application to access 10th answer scripts; verification, revaluation for Class 12 starts on May 31

Latest Question