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 ^ 4 x

Q10  Find the integrals of the functions \sin ^ 4 x

Answers (1)

best_answer

\sin ^ 4 x  can be written as follows using trigonometric identities
\\=\sin^2x.\sin^2x\\ =\frac{1}{4}(1-\cos 2x)^{2}\\ =\frac{1}{4}(1+\cos^22x-2\cos 2x)\\ =\frac{1}{4}(1+\frac{1}{2}(1+\cos 4x)-2\cos 2x)\\ =3/8+\frac{\cos 4x}{8}-\frac{\cos 2x}{2}
 

Therefore, 
\Rightarrow \int \sin^4x\ dx = \int \frac{3}{8}dx+\frac{1}{8}\int\cos 4x\ dx -\frac{1}{2}\int\cos 2x\ dx
                                 = \frac{3x}{8}+\frac{\sin 4x}{32} -\frac{\sin 2x}{4}+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 extends single girl child scholarship registration deadline till November 20
anam-khan

CBSE reminds schools to submit registration data of Class 9, 11 students by October 16

Latest Question