QJEEM20Ma4e02: JEE Main 2020 Mathematics - Class 12 JEEM20Ma4e02 Solution: The integral is equal to:

Question

Verified Solution • 2025–26 Session moderate

The integral

\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx}

is equal to:

Accepted Answer

Given,

I =

\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx}

I = \int\limits_{\pi /6}^{\pi /3} ({2.{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3{\tan ^4}x{\sin ^2}3x.\,2\sin 3xcos\,3x\,)dx

= {1 \over 2}\int\limits_{\pi /6}^{\pi /3} ({4{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3.4{\tan ^4}x{\sin ^3}3xcos\,3x\,)dx

= {1 \over 2}\int\limits_{\pi /6}^{\pi /3} {{d \over {dx}}\left( {{{\tan }^4}x{{\sin }^4}3x} \right)} dx

= {1 \over 2}\left[ {{{\tan }^4}x{{\sin }^4}3x} \right]_{\pi /6}^{\pi /3}

= {1 \over 2}\left[ {9.(0) - {1 \over 3}.{1 \over 3}(1)} \right] = - {1 \over {18}}

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