0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# The value of ∫2nπ0[sinx+cosx]dx, is equal to (where [.] denotes greatest integer function)

A
nπ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
nπ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2nπ
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 A −nπ∫2nπ0[sinx+cosx]dx=∫2nπ0[√2sin(x+π4)]dxThe function is periodic with period 2π.=∫2π0[√2sin(x+π4)]dxHence, first we find the definite integration with limit from 0 to 2πThe graph of sin(x+π4) is shown in the figure. To find values of [√2sin(x+π4)], we can draw the lines y=1√2 and y=−1√2∫2π0[√2sin(x+π4)]dxSplitting the limits for greatest integer function,=∫π201.dx+∫3π4π20.dx+∫π3π4−1.dx+∫3π2π−2.dx+∫7π43π2−1.dx+∫2π7π40.dx=(π2−0)−(π−3π4)−2(3π2−π)−(7π4−3π2)=−πHence, for limit 0 to 2nπ the integral will be n×−π=−nπ

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Property 4
MATHEMATICS
Watch in App
Join BYJU'S Learning Program