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

The value of nππ4π4|sinx+cosx|dx, is equal to

A
22n
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
122n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
none of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 22n
I=nππ4π4|sinx+cosx|dx
=nππ4π42sin(x+π4)dx
=nππ4π42sin(x+π4)dx
sin(x+π4) is periodic with period π
=2n3π4π4sin(x+π4)dx=2n(cos(x+π4))3π4π4=22n

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