Let $\mathrm{n} \geq 2$ be a natural number and $0<\theta<\pi / 2$. Then $\int \frac{\left(\sin ^{\mathrm{n}} \theta-\sin \theta\right)^{\frac{1}{n}} \cos \theta}{\sin ^{n+1} \theta} d \theta$ is equal to: (where C is a constant of integration)

(A) $\frac{n}{n^{2}-1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C$

(B) $\frac{n}{n^{2}+1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C$

(C) $\frac{n}{n^{2}-1}\left(1+\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C$

(D) $\frac{n}{n^{2}-1}\left(1-\frac{1}{\sin ^{n+1} \theta}\right)^{\frac{n+1}{n}}+C$

Indefinite Integrals — JEE Main · 2019 · 10 Jan 2019 · Shift 1 Question with Solution

Let

n \geq 2

be a natural number and

0 < θ < π /2

. Then

\int \frac{( s i n ^{n} θ - s i n θ ) ^{\frac{1}{n}} c o s θ}{s i n ^{n + 1} θ} d θ

is equal to: (where C is a constant of integration)

(A) $\frac{n}{n ^{2} - 1} (1 - \frac{1}{s i n ^{n - 1} θ})^{\frac{n + 1}{n}} + C$ ✓ Correct
(B) $\frac{n}{n ^{2} + 1} (1 - \frac{1}{s i n ^{n - 1} θ})^{\frac{n + 1}{n}} + C$
(C) $\frac{n}{n ^{2} - 1} (1 + \frac{1}{s i n ^{n - 1} θ})^{\frac{n + 1}{n}} + C$
(D) $\frac{n}{n ^{2} - 1} (1 - \frac{1}{s i n ^{n + 1} θ})^{\frac{n + 1}{n}} + C$

Answer: (A)

Substituting

t = sin θ

turns the integral into

\int \frac{( t ^{n} - t ) ^{1/ n}}{t ^{n + 1}} d t

, and factoring

t

out of the bracket as

t (1 - t^{1 - n})

reduces it to

\int \frac{( 1 - t ^{1 - n} ) ^{1/ n}}{t ^{n}} d t

. Now let

z = 1 - t^{1 - n}

, so

d z = (n - 1) t^{- n} d t

and

\frac{d t}{t ^{n}} = \frac{d z}{n - 1}

, leaving

\frac{1}{n - 1} \int z^{1/ n} d z = \frac{1}{n - 1} \cdot \frac{n}{n + 1} z^{(n + 1) / n} + C

. Writing

z

back in terms of

θ

, the result is

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

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option