Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19. 12

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

    Q12.    \int_0^\pi\frac{xdx}{1+\sin x}

Answers (1)

best_answer

We have                   I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}                                           ..........................................................................(i)

By using the identity :-  

                            \ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,   

                                I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin (\Pi -x)}

or                            I\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin x}                                         ............................................................................(ii)

 

Adding both (i) and (ii) we get,     

                        

                               2I\ =\ \int_0^\pi\frac{\Pi}{1+\sin x} dx

or                             2I\ =\ \Pi \int_0^\pi\frac{1-\sin x}{(1+\sin x)(1-\sin x)} dx\ =\ \Pi \int_0^\pi\frac{1-\sin x}{\cos^2 x} dx

or                             2I\ =\ \Pi \int_0^\pi (\sec^2\ -\ \tan x \sec x) x dx

or                               I\ =\ \Pi

 

 

Posted by

Devendra Khairwa

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question