wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solve:x2sin2xdx

Open in App
Solution

x2sin2xdx=x2sin2xdx[ddxx2.sin2xdx]dx=x22cos2x+x.cos2xdx=x22cos2x+x2sin2xcos2x4+C=2xsin2x2x2cos2xcos2x4+C

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