Question

The integral ${\int }_{\pi /6}^{\pi /3}{\sec }^{2/3}x.co{\sec }^{4/3}xdx$ equal to:

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

Answer 1

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

$I=\int \frac{1}{{\cos }^{2/3}x{\sin }^{1/3}x\sin x}dx=\int \frac{{\tan }^{2/3}x}{{\tan }^{2}x}.{\sec }^{2}xdx=\int \frac{{\sec }^{2}x}{{\tan }^{4/3}x}dx$

$(\tan x=t;{\sec }^{2}xdx=dt)=\int \frac{dt}{t^{4/3}}=\frac{t^{-1/3}}{-1/3}=-3\left(t^{-1/3}\right)$

$\Rightarrow I=-3\tan {\left(x\right)}^{-1/3}\Rightarrow I={\left|\frac{3}{\tan {\left(x\right)}^{-1/3}}\right|}_{\pi /6}^{\pi /3}=-3(\frac{1}{{\left(\sqrt{3}\right)}^{1/3}}-{\left(\sqrt{3}\right)}^{1/3})=3(3^{1/3}-\frac{1}{3^{1/6}})=3^{7/6}-3^{5/6}$