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:

Given, <br><br>I = $$\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $$ <br><br>$$I = \int\limits_{\pi /6}^{\pi /3} ({2.{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3{\tan ^4}x{\sin ^2}3x.\,2\sin 3xcos\,3x\,)dx$$<br><br>$$ = {1 \over 2}\int\limits_{\pi /6}^{\pi /3} ({4{{\tan }^3}} x{\sec ^2}x{\sin ^4}3x + 3.4{\tan ^4}x{\sin ^3}3xcos\,3x\,)dx$$<br><br>$$ = {1 \over 2}\int\limits_{\pi /6}^{\pi /3} {{d \over {dx}}\left( {{{\tan }^4}x{{\sin }^4}3x} \right)} dx$$<br><br>$= {1 \over 2}\left[ {{{\tan }^4}x{{\sin }^4}3x} \right]_{\pi /6}^{\pi /3}$<br><br>$$ = {1 \over 2}\left[ {9.(0) - {1 \over 3}.{1 \over 3}(1)} \right] = - {1 \over {18}}$$