Question:
The integral
$\int_{\pi / 6}^{\pi / 3} \tan ^{3} x \cdot \sin ^{2} 3 x\left(2 \sec ^{2} x \cdot \sin ^{2} 3 x+3 \tan x \cdot \sin 6 x\right) d x$
is equal to:
Correct Option: , 3
Solution:
$\int_{\pi / 6}^{\pi / 3}\left[\frac{1}{2} \frac{d\left(\tan ^{4} x\right)}{d x} \cdot \sin ^{4} 3 x+\frac{1}{2} \tan ^{4} x \cdot \frac{d\left(\sin ^{4} 3 x\right)}{d x}\right] d x$
$=\frac{1}{2} \int_{\pi / 6}^{\pi / 3} d\left(\tan ^{4} x \cdot \sin ^{4} 3 x\right) d x$
$=\left[\frac{\tan ^{4} x \sin ^{4} 3 x}{2}\right]_{\pi / 6}^{\pi / 3}=\frac{9 \cdot 0}{2}-\frac{\frac{1}{9} \cdot 1}{2}=\frac{-1}{18}$