# Computing an integral

Find what is equal $I$ to: $$I=\int_{0}^{\pi/2} \sqrt[3]{\tan x}dx$$

I found this simple looking integral, but I don't know if my logic is right. Let $J$ be the next integral with $x\in(0, \frac{\pi}{2})\Rightarrow$ $$J=\int \sqrt[3]{\cot x}dx$$ Then \begin{align} I\pm J & =\int (\sqrt[3]{\tan x}\pm \sqrt[3]{\cot x})dx \\ & =\int \frac{(\sqrt[3]{\tan x})^2 \pm 1}{\sqrt[3]{\tan x}}dx \\ & =\int \frac{\tan x \pm \sqrt[3]{\tan x}}{\sqrt[3]{\tan x}\sqrt[3]{\tan x}}dx \end{align} And so forth using integration by parts (haven't done yet). But if I compute $I+J$ and $I-J$, how can I get $I$? Is my approach correct until now?

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It's a long road but start by substituting $u=\tan x$, then substitute $w=\sqrt[3]{u}$ then I would do a very very length partial fraction decomposition. After that its a long cycle of substitutions and partial fractions. – Ali Caglayan Dec 29 '13 at 18:15
– lab bhattacharjee Dec 29 '13 at 18:15
– Mhenni Benghorbal Dec 29 '13 at 19:21

$$\int^{\pi / 2}_{0} \tan^{a} x \ dx \ \ (|a| <1 )$$

$$= \int^{\pi / 2}_{0} \frac{ \sin^{a} x}{\cos^{a} x} dx = \int^{\pi / 2}_{0} \sin^{2(\frac{a}{2} + \frac{1}{2})-1} (x) \cos ^{2(-\frac{a}{2} + \frac{1}{2}) -1} (x) \ dx$$

$$= \frac{1}{2} \ B(\frac{a}{2}+ \frac{1}{2}, -\frac{a}{2}+ \frac{1}{2}) = \frac{1}{2} \ \frac{\Gamma(\frac{a+1}{2}) \Gamma(1 - \frac{a+1}{2})}{\Gamma(1)}$$

$$= \frac{1}{2} \ \frac{\pi}{\sin \ (\frac{\pi (a+1)}{2})} = \frac{1}{2} \frac{\pi}{\cos (\frac{\pi a}{2})}$$

Therefore,

$$\int^{\pi / 2}_{0} \tan^{\frac{1}{3}} x \ dx = \frac{1}{2} \frac{\pi}{\cos \left(\frac{\pi}{6} \right)} = \frac{\pi}{\sqrt{3}}$$

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