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

sinx(3+2cosx)dx

Open in App
Solution

sinx(3+2cosx)dx
(3sinx+2sinxcosx)dx
3sinx+sin2xdx [2sinxcosx=sin2x]
=3(cosx)+(cos2x)2+c
=3cosxcos2x2+c
Hence, the answer is 3cosxcos2x2+c.

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