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

secxsin(2x+α)+sinαdx is equal to

A
2secα(tanx+tanα)+c
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2secα(tanxtanα)+c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2secα(tanαtanx)+c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2tanα(secxsecα)+c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2secα(tanx+tanα)+c
secxsin(2x+α)+sinαdx
=secxsin2xcosα+cos2xsinα+sinαdx
=secxsin2xcosα+(cos2x+1)sinαdx
=secx2cosx(sinxcosα+cosxsinα)dx
=secx2cos2xcosα(tanx+tanα)dx
=12cosαsec2x(tanx+tanα)dx
Put, tanx+tanα=t
Differentiating above equation w.r.t. x,we get
sec2xdx=dt
=12cosαdtt
=12cosαt1/212+C
=2cosαtanx+tanα+C

2secα(tanx+tanα)+c

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