Solve this

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 :

  1. $\frac{9}{2}$

  2. $-\frac{1}{9}$

  3. $-\frac{1}{18}$

  4. $\frac{7}{18}$


Correct Option: , 3

Solution:

$I=\int_{\pi / 6}^{\pi / 3}\left(\left(2 \tan ^{3} x \cdot \sec ^{2} x \cdot \sin ^{4} 3 x\right)+\left(3 \tan ^{4} x \cdot \sin ^{3} 3 x \cdot \cos 3 x\right)\right) d x$

$\Rightarrow I=\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d\left((\sin 3 x)^{4}(\tan x)^{4}\right)$

$\Rightarrow \mathrm{I}=\left((\sin 3 \mathrm{x})^{4}(\tan \mathrm{x})^{4}\right)_{\pi / 6}^{\pi / 3}$

$\Rightarrow \mathrm{I}=-\frac{1}{18}$

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