JEE Main 2019 — Math solution | Fundamenthol

JEE Main2019Math

Q.

Let $n ⩾ 2$ be a natural number and $0 < θ < π /2$ .

Then $\int \frac{( sin ^{n} θ - sin θ ) ^{\frac{1}{n}} cos θ}{sin ^{n + 1} θ} d θ$ is equal to :

(Where C is a constant of integration)

A
$\frac{n}{n ^{2} - 1} (1 - \frac{1}{sin ^{n + 1} θ})^{\frac{n + 1}{n}} + C$
B
$\frac{n}{n ^{2} + 1} (1 - \frac{1}{sin ^{n - 1} θ})^{\frac{n + 1}{n}} + C$
C
$\frac{n}{n ^{2} - 1} (1 - \frac{1}{sin ^{n - 1} θ})^{\frac{n + 1}{n}} + C$
D
$\frac{n}{n ^{2} - 1} (1 + \frac{1}{sin ^{n - 1} θ})^{\frac{n + 1}{n}} + C$

Solution

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