# Indefinite Integral $\int\sqrt[3]{\tan(x)}dx$

For calculating $\int\sqrt{\tan(x)}dx$, I used this easy method \begin{align}\int\sqrt{\tan(x)}dx&=\frac{1}{2}\int\left(\sqrt{\tan(x)}+\sqrt{\cot(x)}\right)dx+\frac{1}{2}\int\left(\sqrt{\tan(x)}-\sqrt{\cot(x)}\right)dx\\&=\frac{1}{2}\int\frac{\sin(x)+\cos(x)}{\sqrt{\sin(x)\cos(x)}}dx-\frac{1}{2}\int\frac{\cos(x)-\sin(x)}{\sqrt{\sin(x)\cos(x)}}dx\\&=\frac{\sqrt{2}}{2}\int\frac{du}{\sqrt{1-u^2}}-\frac{\sqrt{2}}{2}\int\frac{dv}{\sqrt{v^2-1}}.\end{align}$$u=\sin(x)-\cos(x), v=\sin(x)+\cos(x)$$

Does there exist an easy method for $\int\sqrt[3]{\tan(x)}dx$?

• $\int \sqrt{\tanh (x)} \, dx = \tanh ^{-1}\left(\sqrt{\tanh (x)}\right)-\tan ^{-1}\left(\sqrt{\tanh (x)}\right)$ – alfC Sep 25 '13 at 7:24
• Here's something along the same lines, at least for the cube root. math.stackexchange.com/questions/479865/… – Ron Gordon Sep 25 '13 at 11:35

If you assume $\tan(x)=u^3$, then

$$\int (\tan(x))^{1/3}dx = 3\,\int \!{\frac {{u}^{3}}{{u}^{6}+1}}{du}.$$

For the other one, you can assume $\tan(x)=u^4$ to get

$$\int (\tan(x))^{1/4}dx = 4\,\int \!{\frac {{u}^{4}}{{u}^{8}+1}}{du}.$$

Now, you can use some integration techniques to evaluate the integrals. Note that, for the integral you already did, you can assume $\tan(x)=u^2$ to get

$$= \int (\tan(x))^{1/2} dx = 2\,\int \!{\frac {{u}^{2}}{{u}^{4}+1}}{du}.$$

Note: When you use these substitutions you need the identity

$$\sec^2(x) = 1+\tan^2(x).$$

• In fact OP want to ask whether $\int\dfrac{u^3}{u^6+1}~du$ and $\int\dfrac{u^4}{u^8+1}~du$ have some trickly approaches beside directly taking partial functions. Given that for example $\int\dfrac{u^2}{u^4+1}~du$ has (for example in math.stackexchange.com/questions/425603). – Harry Peter Mar 19 '17 at 8:05