int_{pi / 6}^{pi / 3} tan ^{3} x cdot sin ^{2 | vedclass

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(C) Let $I = \int_{\pi / 6}^{\pi / 3} 	an ^{3} x \cdot \sin ^{2} 3 x\left(2 \sec ^{2} x \cdot \sin ^{2} 3 x+3 	an x \cdot \sin 6 xight) d x$. Note

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$-\frac{1}{18}$

Vedclass · Answer

(C) Let $I = \int_{\pi / 6}^{\pi / 3} 	an ^{3} x \cdot \sin ^{2} 3 x\left(2 \sec ^{2} x \cdot \sin ^{2} 3 x+3 	an x \cdot \sin 6 xight) d x$. Note that $\sin 6x = 2 \sin 3x \cos 3x$. The integrand can be written as: $f(x) = 2 	an^3 x \sec^2 x \sin^4 3x + 3 	an^4 x \sin^2 3x (2 \sin 3x \cos 3x) = 2 	an^3 x \sec^2 x \sin^4 3x + 6 	an^4 x \sin^3 3x \cos 3x$. Observe that $\frac{d}{dx} [(	an x)^4 (\sin 3x)^4] = 4 	an^3 x \sec^2 x \sin^4 3x + 4 	an^4 x \sin^3 3x (3 \cos 3x) = 4 [	an^3 x \sec^2 x \sin^4 3x + 3 	an^4 x \sin^3 3x \cos 3x]$. Thus,the integrand is $\frac{1}{2} \frac{d}{dx} [(	an x)^4 (\sin 3x)^4]$. $I = \frac{1}{2} [	an^4 x \sin^4 3x]_{\pi/6}^{\pi/3} = \frac{1}{2} [(	an^4 \frac{\pi}{3} \sin^4 \pi) - (	an^4 \frac{\pi}{6} \sin^4 \frac{\pi}{2})]$. Since $\sin \pi = 0$ and $	an \frac{\pi}{6} = \frac{1}{\sqrt{3}}$,$\sin \frac{\pi}{2} = 1$: $I = \frac{1}{2} [0 - ((\frac{1}{\sqrt{3}})^4 \cdot 1^4)] = \frac{1}{2} [0 - \frac{1}{9}] = -\frac{1}{18}$.

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

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