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:

Question

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:

$ - {1 \over {9}}$

$ - {1 \over {18}}$

$ {7 \over {18}}$

${9 \over 2}$

Solution

Please login to view the detailed solution steps...

Go to DASH