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

Let n2 be a natural number and 0<θ<π2.
Then (sinnθsinθ)1ncosθsinn+1θdθ is equal to :
(where C is constant of integration)

A
nn21(11sinn1θ)n+1n+C
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
nn2+1(11sinn1θ)n+1n+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
nn21(11sinn+1θ)n+1n+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
nn21(1+1sinn1θ)n+1n+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A nn21(11sinn1θ)n+1n+C
I=(sinnθsinθ)1ncosθsinn+1θdθ
=sinθ(1sin1nθ)1ncosθsinn+1θdθ
=(1sin1nθ)1ncosθsinnθdθ
Take 1sin1nθ=t
(1n)sinnθcosθ dθ=dt
I=1n1t1ndt
=1n1t1n+11n+1+C
=nn21(1sin1nθ)n+1n+C
=nn21(11sinn1θ)n+1n+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