Question:
$\int_{\pi / 6}^{\pi / 3} \tan ^{3} x \cdot \sin ^{2} 3 x\left(2 \sec ^{2} x \cdot \sin ^{2} 3 x+3 \tan x \cdot \sin 6 x\right) d x$
is equal to :
Correct Option: , 3
Solution:
$I=\int_{\pi / 6}^{\pi / 3}\left(\left(2 \tan ^{3} x \cdot \sec ^{2} x \cdot \sin ^{4} 3 x\right)+\left(3 \tan ^{4} x \cdot \sin ^{3} 3 x \cdot \cos 3 x\right)\right) d x$
$\Rightarrow I=\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d\left((\sin 3 x)^{4}(\tan x)^{4}\right)$
$\Rightarrow \mathrm{I}=\left((\sin 3 \mathrm{x})^{4}(\tan \mathrm{x})^{4}\right)_{\pi / 6}^{\pi / 3}$
$\Rightarrow \mathrm{I}=-\frac{1}{18}$