Question

The integral $\underset{\pi /6}{\overset{\pi /3}{\int }}{\sec }^{2/3}x{cosec}^{4/3}xdx$ is equal to :

• A. $3^{5/3}-3^{1/3}$
• B. $3^{4/3}-3^{1/3}$
• C. $3^{5/6}-3^{2/3}$
• D. $3^{7/6}-3^{5/6}$

Answer 1

The correct option is D $3^{7/6}-3^{5/6}$

Given,
$I=\underset{\pi /6}{\overset{\pi /3}{\int }}{\sec }^{2/3}x{cosec}^{4/3}xdx$
$=\underset{\pi /6}{\overset{\pi /3}{\int }}\frac{{\sec }^{2}x}{{\tan }^{4/3}x}dx$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$I=\int \frac{dt}{t^{4/3}}$
$=-3\left(t^{-1/3}\right)$
$={(-3(\tan x{)}^{-\frac{1}{3}})}_{\pi /6}^{\pi /3}$
$=3(3^{1/6}-\frac{1}{3^{1/6}})$
$=3^{7/6}-3^{5/6}$