Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the integrals of the functions sin ^3 ( 2x+ 1 )

Q 4 Find the integrals of the functions \sin ^ 3 ( 2x +1 )

Answers (1)

best_answer

\int \sin^3(2x+1)dx = \int \sin^2(2x+1).\sin(2x+1)dx

The integral can be written as 

                                     = \int (1-\cos^2(2x+1)).\sin(2x+1)dx
Let 
\\\cos (2x+1) =t\\ \sin (2x+1)dx = -dt/2

                                        \\=\frac{-1}{2}\int (1-t^2)dt\\ =\frac{-1}{2}[t-t^3/3]\\ =\frac{t^3}{6}-\frac{t}{2}

Now,  replace the value of t, we get;

=\frac{\cos^3(2x+1)}{6}-\frac{\cos(2x+1)}{2}+C

Posted by

manish

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 explains 'roll number not found' error on portal amid student concerns
anam-khan

Congress attacks Dharmendra Pradhan over CBSE row, alleges failures in exam and revaluation process
anam-khan

CBSE retains Coempt for Class 12 revaluation amid OSM controversy, shifts data to its own servers: Reports

Latest Question