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

Evaluate the integral
π/201+sinx dx

A
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 2

π201+sinxdx

1+sinxdx=(cosx2+sinx2)dx

=2+(sinx2cosx2)+c

=2(sinx2cosx2)+c

π201+sinxdx=[2+(sinx2cosx2)+c]π20

=[2(sinπ4cosπ4)+c][2(sin0cos0)+c]

=2(0)2(1)=2

π201+sinxdx=2


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