16

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 $\int\sqrt[3]{\tan(x)}~dx$?

user91500
  • 5,606
  • 2
    $ \int \sqrt{\tanh (x)} , dx = \tanh ^{-1}\left(\sqrt{\tanh (x)}\right)-\tan ^{-1}\left(\sqrt{\tanh (x)}\right)$ – alfC Sep 25 '13 at 07:24
  • 3
    Here's something along the same lines, at least for the cube root. http://math.stackexchange.com/questions/479865/integrate-int-frac1-sqrt3-tanxdx/479871#479871 – Ron Gordon Sep 25 '13 at 11:35
  • 2
    Not as easy as you hope $$\int \sqrt[3]{\tan (x)} , dx=\frac{3}{4} \sqrt[3]{\tan (x)} \tan (x) , _2F_1\left(\frac{2}{3},1;\frac{5}{3};-\tan ^2(x)\right)+C$$ where $_2F_1$ is the hypergeometric function defined by $$

    _2F_1(a,b;c;z)=\sum _{k=0}^{\infty } \frac{a_k b_k z^k}{k! c_k}$$

     – Raffaele Dec 08 '20 at 13:31
  • When $\tan x < 0$ you want the real cube-root for your integrand? – GEdgar Aug 26 '22 at 00:49

4 Answers4

12

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

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

user91500
  • 5,606
Mhenni Benghorbal
  • 47,431
  • 3
    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 http://math.stackexchange.com/questions/425603). – Harry Peter Mar 19 '17 at 08:05
  • Partial fractions isn't too bad in this case. Since, e.g., the first integrand is odd, we immediately get three simple constraints on the $6$ unknown coefficients. – Travis Willse Dec 08 '20 at 09:03
1

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)

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

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

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

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

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

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

Harry Peter
  • 7,819
1

$\displaystyle \int\sqrt[3]{tanx}dx$

$\displaystyle tanx=u^{\frac{3}{2}}\Rightarrow (1+tan^{2}x)dx=\frac{3}{2}u^{\frac{1}{2}}du$

$\displaystyle\sqrt[3]{tanx}dx=\frac{3}{2}\frac{udu}{1+u^{3}}$

$\displaystyle\frac{u}{u^{3}+1}=\frac{1}{3}(\frac{u}{u+1}-\frac{u(u-2)}{u^{2}-u+1})$

$\displaystyle\frac{u}{u+1}=1-\frac{1}{u+1}$

$\displaystyle\frac{u(u-2)}{u^{2}-u+1}=1-\frac{2u-1}{2(u^{2}-u+1)}-\frac{3}{2(u^{2}-u+1)}$

$\displaystyle\frac{1}{u^{2}-u+1}=\frac{\frac{4}{3}}{(\frac{2}{\sqrt{3}}(u-1))^{2}+1}$

$\displaystyle\frac{3}{2}\frac{u}{u^{3}+1}=\frac{2u-1}{4(u^{2}-u+1)}+\frac{1}{(\frac{2}{\sqrt{3}}(u-1))^{2}+1}-\frac{1}{2(u+1)}$

$\displaystyle\frac{3}{2}\int\frac{udu}{u^{3}+1}=\frac{1}{4}ln\left| u^{2}-u+1 \right|+\frac{\sqrt{3}}{2}$

$\displaystyle\arctan{(\frac{2}{\sqrt{3}}(u-1))}-\frac{1}{2}ln\left| u+1 \right|+c$

$\displaystyle \int\sqrt[3]{tanx}dx=\frac{1}{4}ln\left|\sqrt[3]{ tan^{4}x}-\sqrt[3]{tan^{2}(x)}+1 \right|+\frac{\sqrt{3}}{2}$

$\displaystyle\arctan{(\frac{2}{\sqrt{3}}(\sqrt[3]{tan^{2}(x)}-1))}-\frac{1} {2}ln\left|\sqrt[3]{tan^{2}(x)}+1 \right|+c$

Bouzari Abdelkader
  • 364
0

Maple says, $$ \int \!(\tan \left( x \right))^{1/3} \,{\rm d}x={\frac {1}{4}\ln \left( \left( \tan \left( x \right) \right) ^{{\frac{4}{3}}}- \left( \tan \left( x \right) \right) ^{{\frac{2}{3}}}+1 \right) }+{ \frac {\sqrt {3}}{2}\arctan \left( {\frac {\sqrt {3}}{3} \left( 2\, \left( \tan \left( x \right) \right) ^{2/3}-1 \right) } \right) }-{ \frac {1}{2}\ln \left( \left( \tan \left( x \right) \right) ^{{ \frac{2}{3}}}+1 \right) } $$ But note: this assumes the principal cube root, so it is real only for $\tan x \ge 0$.

GEdgar
  • 111,679