CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

Evaluate sin3xsin2xdx.

Open in App
Solution

sin3xsin2xdx
=sin3x2sinxcosxdx
I=2sin4xcosxdx

Let t=sinxdt=cosxdx

I=2t4dt

=2(t55)+c , where c is the constant of integration

=2sin5x5+c , where t=sinx

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Property 2
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon