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

Find:
π0(sin2x2cos2x2)dx

Open in App
Solution

π0(sin2x2cos2x2)dx
=π0(cos2x2sin2x2)dx
=π0cos2(x2)dx
=π0cosxdx
=[sinx]π0
=(sinπsin0)
=0

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