CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The value of xsinx1cosxdx is equal to
(where C is integration constant)

A
xcotx2+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
xcotx2+C
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
cotx2+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2cotx2+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B xcotx2+C
xsinx1cosxdx=x1cosxdxsinx1cosxdx=12x cosec2(x2)dx2sin(x2)cos(x2)2sin2(x2)dx⎢ ⎢ ⎢ ⎢1cosx=2sin2(x2)sinx=2sin(x2)cos(x2)⎥ ⎥ ⎥ ⎥=[x2 cosec2(x2)+(cot(x2))]dx

This is of the form [f(x)+xf(x)]dx
where
f(x)=cot(x2);f(x)=12 cosec2(x2)
We know that,
[f(x)+xf(x)]dx=xf(x)+C
Thus,
xsinx1cosxdx=xcot(x2)+C

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