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

Evaluate π0log(1cosx)dx.

Open in App
Solution

π0log(1+cosx)dx=π/20log(1+cosx)dx+π/20log(1+cos(πx))dx
=π/20log(sin2x)dx=2π/20log(sinx)dx

I=π0log(sinx)dx=2π/20log(sin2t)dt=2π/20log(2sintcost)dt
=πlog2+2π/20log(sint)dt+2π/20log(cost)dt=πlog2+2I
I=πlog2
π0log(1+cosx)dx=πlog2


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