The integral
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:

 

  1. (1) $\frac{7}{18}$

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

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

  4. (4) $\frac{9}{2}$


Correct Option: , 3

Solution:

$\int_{\pi / 6}^{\pi / 3}\left[\frac{1}{2} \frac{d\left(\tan ^{4} x\right)}{d x} \cdot \sin ^{4} 3 x+\frac{1}{2} \tan ^{4} x \cdot \frac{d\left(\sin ^{4} 3 x\right)}{d x}\right] d x$

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

$=\left[\frac{\tan ^{4} x \sin ^{4} 3 x}{2}\right]_{\pi / 6}^{\pi / 3}=\frac{9 \cdot 0}{2}-\frac{\frac{1}{9} \cdot 1}{2}=\frac{-1}{18}$

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