The integral $\int_{\pi / 6}^{\pi / 3} \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x$ equal to
Correct Option: 1
$I=\int \frac{1}{\cos ^{2 / 3} x \sin ^{1 / 3} x \cdot \sin x} d x$
$=\int \frac{\tan ^{2 / 3} x}{\tan ^{2} x} \cdot \sec ^{2} x \cdot d x$
$=\int \frac{\sec ^{2} x}{\tan ^{4 / 3} x} \cdot d x \quad\left\{\tan x=t, \sec ^{2} x d x=d t\right\}$
$=\int \frac{d t}{t^{4 / 3}}=\frac{t^{-1 / 3}}{-1 / 3}=-3\left(t^{-1 / 3}\right)$
$\Rightarrow I=-3 \tan (x)^{-1 / 3}$
$\Rightarrow I=\left.\frac{3}{(\tan x)^{1 / 3}}\right|_{\pi / 6} ^{\pi / 3}=-3\left(\frac{1}{(\sqrt{3})^{1 / 3}}-(\sqrt{3})^{1 / 3}\right)$
$=3\left(3^{1 / 3}-\frac{1}{3^{1 / 6}}\right)=3^{7 / 6}-3^{5 / 6}$
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