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

(sin4xcos4x)dx

Open in App
Solution

Consider the given integral.

I=(sin4xcos4x)dx

I=((sin2x)2(cos2x)2)dx

I=(sin2xcos2x)(sin2x+cos2x)dx

I=(sin2xcos2x)×1dx

I=(cos2xsin2x)dx

I=cos2xdx

I=[sin2x2]+C


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