CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

Evaluate π0[sin2π2Cos2x2]dx.

Open in App
Solution

π0[sin2π2cosx2]dx
=sin2π2dxcos2x2dx
1cos2π/22dx1+cos2π/22dx
=12cosπ2dx12+cosπ2dx
=x2sinπ2x2+sinπ4
=sinx4sinπ4+C

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