Integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$ 
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$

$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}[\ln(\tan \ x)^{\frac{2}{3}}]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}-\ln(1)^\frac{2}{3}]=\frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}] = \bf \color{red}{\ln(\sqrt{3})}$$ 
However, the answer given is not this value, but instead $= \frac{3\sqrt[3]{3}-3}{2} \approx 0.663... $ while $\ln(\sqrt{3}) \approx 0.549...$
 A: You have $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sec^2 x}{\sqrt[3]{\tan x}}dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sqrt[3]{\tan x}}d\tan x$$
Replacing $t = \tan x$, you have the integration equals
$$\int_{1}^{\sqrt{3}} \frac{1}{\sqrt[3]{t}}dt = \frac{3}{2}t^{\frac{2}{3}} \mid_{1}^{\sqrt{3}} = \frac{3}{2}(\sqrt[3]{3}-1)$$
A: Put $u = \tan x\implies I = \displaystyle \int_{1}^{\sqrt{3}} u^{-\frac{1}{3}}du= \left[\frac{3}{2}u^{\frac{2}{3}}\right]|_{u=1}^{u = \sqrt{3}}= \dfrac{3}{2}\left(\sqrt[3]{3}-1\right)$
A: 
$$
 \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx =
 \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$

This step is wrong since
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x  dx$$
Then $$\frac{f'(x)}{f(x)} =  \frac{2}{3} \frac{ \sec^2x}{\tan x} \neq \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx  dx$$
Try to use trigonometric identities to simplificate it first and then use substitute:
Let $u = \sqrt[3] {\tan{x}}$, then $u^3 = \tan x $, $2u^2 du = \sec^2 x dx$.
So 
$$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{1}^{ 3^{1/6}}\frac{2}{3}\frac{2u^2}{u} du$$
A: Your solution was fine. Just needed to have put $$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}f'(x)dx$$ instead of $$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$which makes your solution $$\frac{3}{2}(f(\frac{\pi}{3})-f(\frac{\pi}{4}))$$where f(x) is the function you defined, namely $$f(x) = (\tan \ x)^{\frac{2}{3}}$$This indeed yields the correct answer,
$$\frac{3}{2}(3^{\frac{1}{3}}-1)$$ 
