If $\int \frac{\cos x d x}{\sin ^{3} x\left(1+\sin ^{6} x\right)^{2/3}}=f(x)\left(1+\sin ^{6} x\right)^{1 / \lambda}+c$ where $c$ is a constant of integration, then $\lambda \mathrm{f}\left(\frac{\pi}{3}\right)$ is equal to:

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Indefinite Integrals

\int \frac{c o s x d x}{s i n ^{3} x ( 1 + s i n ^{6} x ) ^{2/3}} = f (x) (1 + sin^{6} x)^{1/ λ} + c

where

c

is a constant of integration, then

λ f (\frac{π}{3})

is equal to:

Answer: (A)

Substituting

t = sin x

, so that

cos x d x = d t

, turns the integral into

\int \frac{d t}{t ^{3} ( 1 + t ^{6} ) ^{2/3}}

, which rearranges to

\int \frac{d t}{t ^{7} ( 1 + t ^{- 6} ) ^{2/3}}

. Letting

u = 1 + t^{- 6}

gives

d u = - 6 t^{- 7} d t

, so the integral becomes

- \frac{1}{6} \int u^{- 2/3} d u = - \frac{1}{2} u^{1/3} + c = - \frac{1}{2 sin ^{2} x} (1 + sin^{6} x)^{1/3} + c

. Comparing with

f (x) (1 + sin^{6} x)^{1/ λ}

gives

f (x) = - \frac{1}{2} csc^{2} x

and

λ = 3

, so

λ f (\frac{π}{3}) = 3 (- \frac{1}{2} csc^{2} \frac{π}{3}) = 3 (- \frac{1}{2} \cdot \frac{4}{3}) = - 2

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option