JEE Math Practice 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:
- A.$$ - {1 \over {9}}$$
- B.$$ - {1 \over {18}}$$
- C.$$ {7 \over {18}}$$
- D.$${9 \over 2}$$
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