The integral _/6^/3^2/3x^4/3x,dx is equal to:

Question

The integral $\displaystyle\int_{\pi/6}^{\pi/3}\sec^{2/3}x\csc^{4/3}x\,dx$ is equal to:

Accepted Answer

$$I=\int_{\pi/6}^{\pi/3}\frac{\sec^2x}{\tan^{4/3}x}dx$$

Let $\tan x=u\Rightarrow\sec^2x\,dx=du$; limits $u:\dfrac1{\sqrt3}\to\sqrt3$:
$$I=\int_{1/\sqrt3}^{\sqrt3}u^{-4/3}du=\left[-3u^{-1/3}\right]_{1/\sqrt3}^{\sqrt3}=-3\left(3^{-1/6}-3^{1/6}\right)=3^{7/6}-3^{5/6}$$

Solutions · 1