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

Solution 1:

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

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

Solution 2:

For $\int\sqrt[3]{\tan x}~dx$ ,

Let $u=\sqrt[3]{\tan x}$ ,

Then $x=\tan^{-1}u^3$

$dx=\dfrac{3u^2}{u^6+1}~du$

$\therefore\int\sqrt[3]{\tan x}~dx$

$=\int\dfrac{3u^3}{u^6+1}~du$

$=\dfrac{3}{2}\int\dfrac{u^2}{u^6+1}~d(u^2)$

$=\dfrac{3}{2}\int\dfrac{v}{v^3+1}~dv$ (Let $v=u^2$)

With reference to Finding $\int \frac{1}{1+x^3}dx$ without partial fractions,

Let $v=-\dfrac{w-1}{w+1}$ ,

Then $dv=-\dfrac{2}{(w+1)^2}dw$

$\therefore\dfrac{3}{2}\int\dfrac{v}{v^3+1}~dv$

$=\dfrac{3}{2}\int\dfrac{-\dfrac{w-1}{w+1}}{-\dfrac{(w-1)^3}{(w+1)^3}+1}\left(-\dfrac{2}{(w+1)^2}\right)~dw$

$=3\int\dfrac{w-1}{(w+1)^3-(w-1)^3}~dw$

$=\int\dfrac{3(w-1)}{2(3w^2+1)}~dw$

$=\int\dfrac{3w}{2(3w^2+1)}~dw-\int\dfrac{3}{2(3w^2+1)}~dw$

$=\dfrac{\ln(3w^2+1)}{4}-2\sqrt3\tan^{-1}(\sqrt3w)+C$

$=\dfrac{1}{4}\ln\dfrac{3(v-1)^2+(v+1)^2}{(v+1)^2}+2\sqrt3\tan^{-1}\dfrac{\sqrt3(v-1)}{v+1}+C$

$=\dfrac{1}{4}\ln\dfrac{3(u^2-1)^2+(u^2+1)^2}{(u^2+1)^2}+2\sqrt3\tan^{-1}\dfrac{\sqrt3(u^2-1)}{u^2+1}+C$

$=\dfrac{1}{4}\ln\dfrac{3(\tan^\frac{3}{2}x-1)^2+(\tan^\frac{3}{2}x+1)^2}{(\tan^\frac{3}{2}x+1)^2}+2\sqrt3\tan^{-1}\dfrac{\sqrt3(\tan^\frac{3}{2}x-1)}{\tan^\frac{3}{2}x+1}+C$

For $\int\sqrt[4]{\tan x}~dx$ ,

Let $t=\sqrt[4]{\tan x}$ ,

Then $x=\tan^{-1}t^4$

$dx=\dfrac{4t^3}{t^8+1}~dt$

$\therefore\int\sqrt[4]{\tan x}~dx$

$=\int\dfrac{4t^4}{t^8+1}~dt$

$=\int\dfrac{4t^4}{(t^4-\sqrt2t^2+1)(t^4+\sqrt2t^2+1)}~dt$

$=\int\dfrac{\sqrt2t^2}{t^4-\sqrt2t^2+1}~dt-\int\dfrac{\sqrt2t^2}{t^4+\sqrt2t^2+1}~dt$ (with reference to https://www.wolframalpha.com/input/?i=4x%5E2%2F%28x%5E4%2B1%29)

$=\int\dfrac{\sqrt2}{t^2-\sqrt2+\dfrac{1}{t^2}}~dt-\int\dfrac{\sqrt2}{t^2+\sqrt2+\dfrac{1}{t^2}}~dt$

$=\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}-\sqrt2}~dt+\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}-\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}+\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}+\sqrt2}~dt$

$=\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2-\sqrt2}~dt+\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{\left(t+\dfrac{1}{t}\right)^2-2-\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2+\sqrt2}~dt-\dfrac{1}{\sqrt2}\int\dfrac{1-\dfrac{1}{t^2}}{\left(t+\dfrac{1}{t}\right)^2-2+\sqrt2}~dt$

$=\dfrac{1}{\sqrt2}\int\dfrac{d\left(t-\dfrac{1}{t}\right)}{\left(t-\dfrac{1}{t}\right)^2+2-\sqrt2}+\dfrac{1}{\sqrt2}\int\dfrac{d\left(t+\dfrac{1}{t}\right)}{\left(t+\dfrac{1}{t}\right)^2-2-\sqrt2}-\dfrac{1}{\sqrt2}\int\dfrac{d\left(t-\dfrac{1}{t}\right)}{\left(t-\dfrac{1}{t}\right)^2+2+\sqrt2}-\dfrac{1}{\sqrt2}\int\dfrac{d\left(t+\dfrac{1}{t}\right)}{\left(t+\dfrac{1}{t}\right)^2-2+\sqrt2}$

$=\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tan^{-1}\dfrac{t-\dfrac{1}{t}}{2-\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tanh^{-1}\dfrac{t+\dfrac{1}{t}}{2+\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tan^{-1}\dfrac{t-\dfrac{1}{t}}{2+\sqrt2}+\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tanh^{-1}\dfrac{t+\dfrac{1}{t}}{2-\sqrt2}+C$

$=\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tan^{-1}\dfrac{\sqrt[4]{\tan x}-\dfrac{1}{\sqrt[4]{\tan x}}}{2-\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tanh^{-1}\dfrac{\sqrt[4]{\tan x}+\dfrac{1}{\sqrt[4]{\tan x}}}{2+\sqrt2}-\dfrac{1}{\sqrt2\sqrt{2+\sqrt2}}\tan^{-1}\dfrac{\sqrt[4]{\tan x}-\dfrac{1}{\sqrt[4]{\tan x}}}{2+\sqrt2}+\dfrac{1}{\sqrt2\sqrt{2-\sqrt2}}\tanh^{-1}\dfrac{\sqrt[4]{\tan x}+\dfrac{1}{\sqrt[4]{\tan x}}}{2-\sqrt2}+C$