Question

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

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

Answer 1

Answer: (C)

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