Question

$\underset{\pi /6}{\overset{\pi /3}{\int }}{\tan }^{3}x\cdot {\sin }^{2}3x\left(2{\sec }^{2}x\cdot {\sin }^{2}3x+3\tan x\cdot \sin 6x\right)dx$ is equal to :

• A. $\frac{9}{2}$
• B. $-\frac{1}{9}$
• C. $-\frac{1}{18}$
• D. $\frac{7}{18}$

Answer 1

Answer: (C)

$I=\underset{\pi /6}{\overset{\pi /3}{\int }}\left(\left(2{\tan }^{3}x\cdot {\sec }^{2}x\cdot {\sin }^{4}3x\right)+\left(3{\tan }^{4}x\cdot {\sin }^{3}3x\cdot \cos 3x\right)\right)dx$
$\Rightarrow I=\frac{1}{2}{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}d\left((\sin 3x{)}^4(\tan x{)}^4\right)$
$\Rightarrow I={\left((\sin 3x{)}^4(\tan x{)}^4\right)}_{\pi /6}^{\pi /3}$
$\Rightarrow I=-\frac{1}{18}$