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

π/2π/2cos22x1+25xdx=______

A
π4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
π2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
π2
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 A π4
I=π/2π/2cos22x1+25xdx

I=π/2π/2cos22x1+25xdx
Adding these two equations
I=12π/2π/2(cos22x1+25x+cos22x1+25x)dx

I=π/20(cos22x1+25x+cos22x1+25x)dx

I=π/20(cos22x1+25x+25xcos22x1+25x)dx
I=π/20cos22x dx
I=π/201+cos4x2dx
=12[x+sin4x4]π/20

=12[π2+00+0]

=π4

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Property 1
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon