Integral $\displaystyle\int \frac{1}{\sqrt[4]{1-x^4}}dx$ Solution

[juantheron]

$\displaystyle \int\frac{1}{\sqrt[4]{1-x^4}}dx$

 

[CaptainBlack]

$\displaystyle \int\frac{1}{\sqrt[4]{1-x^4}}dx$

This does not have an integral in terms of elementary functions by the theorem of Chebyshev that you have had pointed at in an earlier post.

It can also be found >>here<< near expression (20)

CB

 

[Jameson]

This does not have an integral in terms of elementary functions by the theorem of Chebyshev that you have had pointed at in an earlier post.

It can also be found >>here<< near expression (20)

CB

A quicker way to Wolfram straight from MHB is using one of our MHB Widgets.

View attachment 411

For your question this will use the same Wolfram indefinite integral calculator. Here's the output of your question.

View attachment 410

 

[Opalg]

There is a neat formula (discovered by Gauss) for the definite integral of $1/ \sqrt[4]{1-x^4}$, namely $$\int_0^1 \frac{1}{\sqrt[4]{1-x^4}}dx = \frac{\pi}{2M(\sqrt2)},$$ where $M(x)$ denotes the arithmetic-geometric mean of 1 and $x$. See the very interesting article An Eloquent Formula for the Perimeter of an Ellipse in last month's issue of the Notices of the American Math. Soc.

 

[chisigma]

There is a neat formula (discovered by Gauss) for the definite integral of $1/ \sqrt[4]{1-x^4}$, namely $$\int_0^1 \frac{1}{\sqrt[4]{1-x^4}}dx = \frac{\pi}{2M(\sqrt2)},$$ where $M(x)$ denotes the arithmetic-geometric mean of 1 and $x$. See the very interesting article An Eloquent Formula for the Perimeter of an Ellipse in last month's issue of the Notices of the American Math. Soc.

According to...

Gauss's Constant -- from Wolfram MathWorld

... the so called 'Gauss's constant' is defined as... $\displaystyle G= \frac{1}{\text{M}\ (1, \sqrt{2})} = \frac{2}{\pi} \int_{0}^{1} \frac{dx}{\sqrt{1-x^{4}}}$

Kind regards

$\chi$ $\sigma$

 

[juantheron]

But I have solved like that way.

(Dont no whether it is Right or not)

$\displaystyle \int\frac{1}{\sqrt[4]{1-x^4}}dx$

Put $\displaystyle 1-x^4=t^4x^4\Leftrightarrow x^4 = \frac{1}{1+t^4}$

$\displaystyle4x^3 dx = -\frac{4t^3}{(1+t^4)^2}dt$

$\displaystyle dx = -\frac{t^3}{(1+t^4)^2.x^3}dt$

So Integral Convert into $\displaystyle -\int\frac{t^3}{(1+t^4)^2.x^3.tx}$

$\displaystyle - \int\frac{t^2}{(1+t^4)^2}.\frac{(1+t^4)}{1}dt$

$\displaystyle -\int\frac{t^2}{1+t^4}dt = -\frac{1}{2}\int\frac{(t^2+1)+(t^2-1)}{1+t^4}dt$

Now Let $\displaystyle \mathbb{I = \int \frac{t^2+1}{t^4+1}}$ and $\displaystyle\mathbb{J=\int\frac{t^2-1}{t^4+1}}dt$

after that we can solve for $\mathbb{I}$ and $\mathbb{J}$

Thanks

 

[chisigma]

But I have solved like that way.

(Dont no whether it is Right or not)

$\displaystyle \int\frac{1}{\sqrt[4]{1-x^4}}dx$

Put $\displaystyle 1-x^4=t^4x^4\Leftrightarrow x^4 = \frac{1}{1+t^4}$

$\displaystyle4x^3 dx = -\frac{4t^3}{(1+t^4)^2}dt$

$\displaystyle dx = -\frac{t^3}{(1+t^4)^2.x^3}dt$

So Integral Convert into $\displaystyle -\int\frac{t^3}{(1+t^4)^2.x^3.tx}$

$\displaystyle - \int\frac{t^2}{(1+t^4)^2}.\frac{(1+t^4)}{1}dt$

$\displaystyle -\int\frac{t^2}{1+t^4}dt = -\frac{1}{2}\int\frac{(t^2+1)+(t^2-1)}{1+t^4}dt$

Now Let $\displaystyle \mathbb{I = \int \frac{t^2+1}{t^4+1}}$ and $\displaystyle\mathbb{J=\int\frac{t^2-1}{t^4+1}}dt$

after that we can solve for $\mathbb{I}$ and $\mathbb{J}$

Thanks

Very good!... pratically speacking You can directly solve the integral...

$\displaystyle - \int\frac{t^2}{1+t^4}dt = \frac{1}{4\ \sqrt{2}}\ [\ln (t^{2}+ \sqrt{2}\ t +1) - \ln (t^{2} - \sqrt{2}\ t +1) + 2\ \tan^{-1} (1+\sqrt{2}\ t) - 2\ \tan^{-1} (1- \sqrt{2}\ t)] + c$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

This does not have an integral in terms of elementary functions by the theorem of Chebyshev that you have had pointed at in an earlier post.

It can also be found >>here<< near expression (20)

CB

From...

Indefinite Integral -- from Wolfram MathWorld

Chebyshev proved that if $u$,$v$ and $w$ are rational numbers, then... $\displaystyle \int x^{u}\ (a + b\ x^{v})^{w}$ (1)...is integrable in terms of elementary functions iff $\frac{u+1}{v}$, $w$ or $w+\frac{u+1}{v}$ is an integer... In this case is $w=-\frac{1}{4}$, $u=0$ and $v=4$ so that... $\displaystyle w+\frac{u+1}{v}=0$ (2)

... and, if I remember well, $0 \in \mathbb{Z}$...

Kind regards

$\chi$ $\sigma$

 

[juantheron]

To Chisigma

How can I calculate it Directily

$\displaystyle - \int\frac{t^2}{1+t^4}dt = \frac{1}{4\ \sqrt{2}}\ [\ln (t^{2}+ \sqrt{2}\ t +1) - \ln (t^{2} - \sqrt{2}\ t +1) + 2\ \tan^{-1} (1+\sqrt{2}\ t) - 2\ \tan^{-1} (1- \sqrt{2}\ t)] + c$

Thanks

 

[chisigma]

To Chisigma

How can I calculate it Directily

$\displaystyle - \int\frac{t^2}{1+t^4}dt = \frac{1}{4\ \sqrt{2}}\ [\ln (t^{2}+ \sqrt{2}\ t +1) - \ln (t^{2} - \sqrt{2}\ t +1) + 2\ \tan^{-1} (1+\sqrt{2}\ t) - 2\ \tan^{-1} (1- \sqrt{2}\ t)] + c$

Thanks

You can expand the function under integral sign in partial fractions...

$\displaystyle f(t)=\frac{t^{2}}{1+t^{4}}= \frac{r_{1}}{t-\sqrt{2}\ (1+i)} + \frac{r_{2}}{t-\sqrt{2}\ (1-i)} + \frac{r_{3}}{t+\sqrt{2}\ (1+i)} + \frac{r_{4}}{t+\sqrt{2}\ (1-i)}$ (1)

... where...

$\displaystyle r_{1}= \lim_{t \rightarrow \sqrt{2}\ (1+i)} (t-\sqrt{2}\ (1+i))\ f(t)$

$\displaystyle r_{2}= \lim_{t \rightarrow \sqrt{2}\ (1-i)} (t-\sqrt{2}\ (1-i))\ f(t)$

$\displaystyle r_{3}= \lim_{t \rightarrow - \sqrt{2}\ (1+i)} (t+\sqrt{2}\ (1+i))\ f(t)$

$\displaystyle r_{4}= \lim_{t \rightarrow -\sqrt{2}\ (1-i)} (t+\sqrt{2}\ (1-i))\ f(t)$ (4)

... and then integrate 'term by term'...

Kind regards

$\chi$ $\sigma$