Let n ≥ 2 be a natural number and 0 < θ < π/2. Then ∫ {(\sin^n \theta-\sin \theta)^{1/n} \cos\theta}/{\sin^{n+1} \theta} dθ is equal to : (Where C is a constant of integration) [10 Jan 2019 Shift 1]

Correct Answer is : nn2−1 (1−1sinn−1θ)n+1n + C

n/{n^2-1} (1-1/{\sin^{n+1} \theta})^{{n+1}/n} + C

n/{n^2+1} (1-1/{\sin^{n-1} \theta})^{{n+1}/n} + C

n/{n^2-1} (1-1/{\sin^{n-1} \theta})^{{n+1}/n} + C

n/{n^2-1} (1+1/{\sin^{n-1} \theta})^{{n+1}/n} + C

JEE Main 10 Jan 2019 Shift 1 Solved Paper

Section: Mathematics

Question : 60 of 90

Marks: +1, -0

Let n ≥ 2 be a natural number and 0 < θ < π/2.
Then ∫

(sinnθ−sinθ)

cosθ

sinn+1θ

dθ is equal to :
(Where C is a constant of integration)

[10 Jan 2019 Shift 1]

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