The integral ∫π/6π/3 tan3⁡xsin2⁡3x2sec2⁡xsin2⁡3x+3tan⁡xsin⁡6xdx is equal to

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

Solution:

Let,

$I = \int_{π / 6}^{π / 3} \tan^{3} x \sin^{2} 3 x (2 \sec^{2} x \sin^{2} 3 x + 3 \tan x \sin 6 x) d x .$

Then,

$\begin{array}{l} I = \int_{π / 6}^{π / 3} \{\frac{d}{d x} (\frac{1}{2} \tan^{4} x) \sin^{4} 3 x + \tan^{4} x \frac{d}{d x} (\frac{1}{2} \sin^{4} 3 x)\} d x \\ \Rightarrow I = \frac{1}{2} \int_{π / 6}^{π / 3} \frac{d}{d x} (\tan^{4} x \sin^{4} 3 x) d x \\ \Rightarrow I = \frac{1}{2} {[\tan^{4} x \sin^{4} 3 x]}_{π / 6}^{π / 3} = \frac{1}{2} (9 \times 0 - \frac{1}{9}) = - \frac{1}{18} \end{array}$

The integral $\int_{π / 6}^{π / 3} \tan^{3} x \sin^{2} 3 x (2 \sec^{2} x \sin^{2} 3 x + 3 \tan x \sin 6 x) d x$ is equal to

Solution:

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