CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

 π/20log(sinx) dx= 


A
(π2)log2
loader
B
πlog12
loader
C
πlog12
loader
D
12log2
loader

Solution

The correct option is B (π2)log2
I=π/20log(sinx) dx=π/20log(cosx) dx2I=π/20log(sinx cosx) dx=π/20log(sin2x) dxπ/20log(2) dx
=12π0log(sint) dtπ2log2     [Putting in first integral 2x=t]=122π/20logsintdtπ2log2       [2a0f(x)dx=2a0f(x)dx if f(2ax)=f(x)]2II=π2log2I=π2log2.      [baf(x)dx=baf(t)dt]  

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image