Evaluation of $\int\sqrt[4]{\tan x}dx$

Question

Evaluation of $\displaystyle \int\sqrt[4]{\tan x}dx$

$\bf{My\; Try}::$ Let $\tan x = t^4\;\;,$ Then $\sec^2 xdx = 4t^3dt$. So $\displaystyle dx = \frac{4t^3}{1+t^8}dt$

So Integral Convert into $\displaystyle 4\int\frac{t^4}{1+t^8}dt = 2\int \frac{(t^4+1)+(t^4-1)}{t^8+1}dt$

So Integral is $\displaystyle 2\int\frac{t^4+1}{t^8+1}dt+2\int\frac{t^4-1}{t^4+1}dt$

Now How can I solve after that

Help me

Thanks

I get this very long and nasty expression from Wolf – S L Jul 08 '14 at 17:10 — S L, Jul 08 '14 at 17:10
Using Wolf amounts to cheating :( . – Count Iblis Jul 08 '14 at 17:14 — Count Iblis, Jul 08 '14 at 17:14
Here is the same problem. – Mhenni Benghorbal Jul 08 '14 at 18:21 — Mhenni Benghorbal, Jul 08 '14 at 18:21

score 3 · Answer 1 · answered Jan 18 '21 at 00:07

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)

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

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

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

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

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

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

Have you checked this? Wolfy chokes on it. – marty cohen Jan 18 '21 at 01:15 — marty cohen, Jan 18 '21 at 01:15

score 3 · Answer 2 · answered Mar 22 '22 at 08:45

Transforming the integral into an integral of rational function by letting $y^{4}=\tan x$, then

$$ I=4 \int \frac{y^{4}}{1+y^{8}} d y $$

By my post,

$$ \displaystyle \int \frac{x^{4}}{1+x^{8}} d x =\frac{1}{4 \sqrt{2}}\left[\frac{1}{2 \sqrt{2+\sqrt{2}}} \ln \left|\frac{x^{2}-\sqrt{2+\sqrt{2}} x+1}{x^{2}+\sqrt{2+\sqrt{2}} x+1}\right|+\frac{1}{\sqrt{2-\sqrt{2}}} \tan^{-1}\left(\frac{x^{2}-1}{\sqrt{2-\sqrt{2}} x}\right)\\ \quad - \frac{1}{2 \sqrt{2-\sqrt{2}}} \ln \left|\frac{x^{2}-\sqrt{2-\sqrt{2}} x+1}{x^{2}+\sqrt{2-\sqrt{2} x}+1}\right|-\frac{1}{\sqrt{2+\sqrt{2}}} \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2+\sqrt{2}} x}\right)\right]+C$$

Now we can conclude that

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

:|D Wish you enjoy the solution!

score 2 · Answer 3 · answered Jul 08 '14 at 17:45

Let's compute $\int\frac{t^4}{t^8+1}dt$ using partial fractions.

We start with locating the poles of the integrand in the complex plane. They are at $t_n = \exp\left[\frac{(2 n+1) i\pi}{8}\right]$. Then near each pole the singular behavior of the integrand is:

$$t_n^4\left[\lim_{t\to t_n}\frac{t-t_n}{t^8 + 1}\right]\frac{1}{t-t_n}=\frac{1}{8 t_n^3}\frac{1}{t-t_n}$$

Then the sum of this over n from n = 0 to n = 7 is the partial fraction expansion (to see this, subtract this from the integrand, the resulting expression doesn't have any singularities anymore therefore it must be a polynomial, but it tends to zero at infinity so it is identical to zero).

To compute the integral, you can integrate this term by term. To convert the expression with complex terms to a manifestly real expression, you can take the terms involving $t_n$ and $t_{8-n}$ together, they are complex conjugates of each other, so you can replace them by twice the real part of the term involving $t_n$ (except for n = 4 in which case the term is real). So, we need to simplify:

$$\Re\frac{1}{4 t_n^3}\log(t-t_n)$$

This is easy, we have an expression of the form

$$\exp(i\alpha)\log(t-\exp(i\beta))$$

The real and imaginary parts of the logarithm are obtained as follows. You write the arguments as the sum of the real and imaginary parts:

$$\log\left[t - \cos(\beta) - i\sin(\beta)\right]$$

And using

$$x+i y= \sqrt{x^2 + y^2}\exp(i\arctan(\frac{y}{x}))$$

where I've assumed for simplicity that $\arctan(\frac{y}{x})$ yields the correct argument of the complex number.

So, obtaining the integral is straightforward using partial fractions.

Quanto · Answer 4 · 2022-03-30T22:33:32.390

1

In general, with $t=\sqrt[n]{\tan x}$

\begin{align} &\int\sqrt[n]{\tan x}\>dx=n\int \frac{t^n}{1+t^{2n}}dt\\ =&\>\frac14 \sum_{k=1}^{2n} (-1)^{k-1}\left(\sin a_k \ln(x^2-2x\cos a_k+1)-2\cos a_k \tan^{-1}\frac{\sin a_k}{x-\cos a_k}\right) +C \end{align} where $a_k=\frac{(2k-1)\pi}{2n}$.

edited Mar 30 '22 at 22:33

answered Mar 30 '22 at 21:51

Quanto

120,125

heropup · Answer 5 · 2014-07-08T17:51:09.390

You are on the right track, but the remaining computations are tedious. First, note that we have the factorizations $$t^4 + 1 = (t^2 - \sqrt{2} t + 1)(t^2 + \sqrt{2} t + 1),$$ and $$t^8 + 1 = (t^2 - \alpha t + 1)(t^2 + \alpha t + 1)(t^2 - \beta t + 1)(t^2 + \beta t + 1),$$ where $\alpha = \sqrt{2+\sqrt{2}}$ and $\beta = \sqrt{2 - \sqrt{2}} = \sqrt{2}/\alpha$. These then admit partial fraction decomposition of the two integrands you obtained, from which we can then eventually arrive at an antiderivative.

Incidentally, the separation into two terms as you have done is not particularly useful.

An alternate approach would be to perform the partial fraction decomposition of $\dfrac{t^4}{t^8+1}$ into linear factors over $\mathbb C$, for which the antiderivative is easier to compute as these will be logarithms of the form $$\log (t - \zeta_{16}^{2k+1})$$ for $\zeta_{16} = e^{\pi i/8}$ a primitive $16^{\rm th}$ root of unity and $k = 0, 1, \ldots, 7$. Then we recombine these appropriately to obtain an expression without complex numbers.

Evaluation of $\int\sqrt[4]{\tan x}dx$

5 Answers5