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

Solve the integral I=π0sin2x dx

A
π
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
π2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
π4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B π2
We have, I=π0sin2xdx
Using trigonometric identity sin2x=1cos2x2
I=π0[1cos2x]2dx
=12[π0dxπ0cos2xdx]
=12[xsin2x2]π0
=12[(π0)(sin2πsin0)2]
I=π2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon