Question

The integral ${\int }_{\pi /6}^{\pi /3}{\tan }^{3}x\cdot {\sin }^{2}3x(2{\sec }^{2}x\cdot {\sin }^{2}3x+3\tan x\cdot \sin 6x)dx$is equal to

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

Answer 1

The correct option is B $-\frac{1}{18}$

${\int }_{\pi /6}^{\pi /3}{\tan }^{3}x\cdot 2{\sec }^{2}x\cdot {\sin }^{4}3x+3{\tan }^{4}x\cdot {\sin }^{2}3x\cdot 2\sin 3x\cdot \cos 3x)dx$
$=\frac{1}{2}{\int }_{\pi /6}^{\pi /3}4{\tan }^{3}x\cdot {\sec }^{2}x\cdot {\sin }^{4}3x+3\cdot 4{\tan }^{4}x\cdot {\sin }^{3}3x\cdot \cos 3x)dx$

$=\frac{1}{2}{\int }_{\pi /6}^{\pi /3}\frac{d}{dx}({\tan }^{4}x\cdot {\sin }^{4}3x)dx$
$=\frac{1}{2}{[{\tan }^{4}x\cdot {\sin }^{4}3x]}_{\pi /6}^{\pi /3}$
$=\frac{1}{2}[9\times (0)-\frac{1}{9}(1{)}^4]$
$=-\frac{1}{18}$