Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
8104911739
For Business Enquiry

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

find integral of (sinx sin2x sin3x)?

Asked by Aparna 16th March 2013, 11:58 AM
Answered by Expert
Answer:
I = ? sin x. sin 2x. sin 3x dx

I= (1/2) ? [ 2 sin 2x. sin x ] sin 3x dx

I= (1/2) ? [ cos ( 2x-x) - cos ( 2x+x) ] sin 3x dx

I= (1/2) ? ( cos x - cos 3x ) sin 3x dx

I= (1/4) ? [ ( 2 sin 3x cos x ) - ( 2 sin 3x cos 3x ) ] dx

I= (1/4) ? { [ sin ( 3x + x ) + sin ( 3x - x ) ] - ( sin 23x ) } dx

I= (1/4) ? [ ( sin 4x + sin 2x ) - ( sin 6x ) ] dx

I= (/4) { [ ( - cos 4x ) / 4 ] + [ ( - cos 2x ) / 2 ] - [ ( - cos 6x ) / 6 ] } + C

I= ( -1/16) cos 4x - (1/8) cos 2x + (1/24) cos 6x + C 
Answered by Expert 16th March 2013, 7:52 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp