The integral $\int_{\pi/6}^{\pi/3}\tan^3 x\cdot\sin^2 3x\left(2\sec^2 x\cdot\sin^2 3x+3\tan x\cdot\sin 6x\right)dx$ is

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Definite Integrals

The integral

\int_{π /6}^{π /3} tan^{3} x \cdot sin^{2} 3 x (2 sec^{2} x \cdot sin^{2} 3 x + 3 tan x \cdot sin 6 x) d x

Answer: (A)

The integrand is an exact derivative:

tan^{3} x sin^{2} 3 x (2 sec^{2} x sin^{2} 3 x + 3 tan x sin 6 x) = \frac{1}{2} \frac{d}{d x} (tan^{4} x sin^{4} 3 x)

. Hence the integral equals

\frac{1}{2} [tan^{4} x sin^{4} 3 x]_{π /6}^{π /3}

. At

x = \frac{π}{3}

the bracket is

(3)^{4} sin^{4} π = 0

, and at

x = \frac{π}{6}

it is

(\frac{1}{3})^{4} sin^{4} \frac{π}{2} = \frac{1}{9}

, so the value is

\frac{1}{2} (0 - \frac{1}{9}) = - \frac{1}{18}

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