The integral \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{ an }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3 an x.\sin 6x} ight)dx} is equal to:

Question

The integral \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{	an }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3	an x.\sin 6x} ight)dx} is equal to:

Accepted Answer

Answer: B. - {1 \over {18}} Given, I = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{	an }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3	an x.\sin 6x} ight)dx} I = \int\limits_{\pi /6}^{\pi /3} ({2.{{	an }^3}} x{\sec ^2}x{\sin ^4}3x + 3{	an ^4}x{\sin ^2}3x.\,2\sin 3xcos\,3x\,)dx = {1 \over 2}\int\limits_{\pi /6}^{\pi /3} ({4{{	an }^3}} x{\sec ^2}x{\sin ^4}3x + 3.4{	an ^4}x{\sin ^3}3xcos\,3x\,)dx = {1 \over 2}\int\limits_{\pi /6}^{\pi /3} {{d \over {dx}}\left( {{{	an }^4}x{{\sin }^4}3x} ight)} dx = {1 \over 2}\left[ {{{	an }^4}x{{\sin }^4}3x} ight]_{\pi /6}^{\pi /3} = {1 \over 2}\left[ {9.(0) - {1 \over 3}.{1 \over 3}(1)} ight] = - {1 \over {18}}

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:

Answer

Explanation

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