Integral $\int_0^1\sqrt{1-x^4}dx$ I am asked to show $\int_0^1\sqrt{1-x^4}dx=\frac{\{\Gamma(1/4)\}^2}{6\sqrt{2\pi}}$. I know the gamma function is defined by $\Gamma(n)=\int_0^\infty x^{n-1}e^{-x}dx$. I tried to substituted $x^2=\sin(t)$ but couldn't go further. I am really questioned how a radical function can convert to an exponential one? :-0 Thank you.
 A: HINT: Change variables $t=x^4$, and use  Euler's integral of the first kind to express the answer as beta function.
A: $\int \sqrt{1-x^4}dx$
Using Integration by Parts, lets say u(x)=$\sqrt{1-x^4}$, v'(x)=1.  The chain rule and exponent rule shows that u'(x)=$-\frac{2x^3}{\sqrt{1-x^4}}$, and v(x) is just x.
$\int \sqrt{1-x^4}dx = x\sqrt{1-x^4}- \int\frac{-2x^4}{\sqrt{1-x^4}}dx$
$\int \sqrt{1-x^4}dx = x\sqrt{1-x^4}-\int \frac{2-2x^4}{\sqrt{1-x^4}}dx+\int \frac{2}{\sqrt{1-x^4}}dx$
$\int \sqrt{1-x^4}dx = x\sqrt{1-x^4}-2\int \sqrt{1-x^4}dx+2\int \frac{1}{\sqrt{1-x^4}}dx$
Now, we add $2\int \sqrt{1-x^4}dx $ to both sides
$3\int \sqrt{1-x^4}dx = x\sqrt{1-x^4}+2\int \frac{1}{\sqrt{1-x^4}}dx$
Now, we divide both sides by 3
$\int \sqrt{1-x^4}dx= \frac{x\sqrt{1-x^4}+2\int \frac{1}{\sqrt{1-x^4}}dx}3$
$\int \sqrt{1-x^4}dx= \frac{x\sqrt{1-x^4}+2\int \frac{1+x²}{\sqrt{1-x^4}}dx-2\int\frac{x²}{\sqrt{1-x^4}}dx}3$
Alright, now let's take a quick sidetrack and solve for $\int \frac{1+x²}{\sqrt{1-x^4}}dx$
$\int\frac{1+x²}{\sqrt{1-x^4}}dx=\int\frac{1+x²}{\sqrt{(1-x²)(x²+1)}}dx=\int\sqrt{\frac{1+x²}{1-x²}}dx$
substitute u=$sin^{(-1)}(x)$
 $\frac{du}{dx}=\frac{1}{\sqrt{1-x²}}$, so $dx=\sqrt{1-x²} du$
$\int \sqrt{1+sin²(u)}du$
This boils down to the incomplete elliptic integral of the second kind.
E(u|-1)+$C$
u=$sin^{(-1)}(x)$
$\int \frac{1+x²}{\sqrt{1-x^4}}dx=E(sin^{(-1)}(x)|-1)+C$
Let's plug it in to the original equation
$\int\sqrt{1-x^4}dx=\frac{x\sqrt{1-x^4}+2E(sin^{(-1)}(x)|-1)-2\int\frac{x²}{\sqrt{1-x^4}}dx}3$
Now, let's solve for $\int\frac{x²}{\sqrt{1-x^4}}dx $
substitute v=$sin^{(-1)}(x²)$, x=$\sqrt{sin(v)}$, dx/dv=$\frac{cos(v)}{2\sqrt{sin(v)}}$
$\int\frac{sin(v)}{\sqrt{1-sin²(v)}}•\frac{cos(v)}{2\sqrt{sin(v)}}dv$
$\int\frac{\sqrt{sin(v)}}{cos(v)}*\frac{cos(v)}2 dv$
$\frac{1}{2}\int\sqrt{sin(v)}dv$
substitute w=$\frac{π}{2}$-v, dv=-dw
$-\frac{1}{2}\int\sqrt{sin(\frac{π}2-w)}dw$
$-\frac{1}{2}\int\sqrt{cos(w)}dw$
$-\frac{1}{2}\int\sqrt{1-2sin²(\frac{w}2)}dw$
E($\frac{w}2$|2)+C
E($\frac{\frac{π}2-v}2$|2)+C
E($\frac{\frac{π}2-sin^{(-1)}(x²)}2$|2)+C
E($\frac{cos^{(-1)}(x²)}2$|2)+C
Plug it back in,
$\int\sqrt{1-x^4}dx= \frac{x\sqrt{1-x^4}+2E(sin^{(-1)}(x)|-1)-2E(\frac{cos^{(-1)}(x²)}2|2)}3$+C
Now, this may look like the final solution, but after plugging it into my calculator, I've seen that there is a slight catch, the value E($\frac{cos^{(-1)}(x²)}2$|2) can be positive or negative.  I've concluded that, as is the case with the derivative of $sec^{(-1)}$(x) it's sign is dependent on the real value of x. So the final solution is
$\int\sqrt{1-x^4}dx= \frac{x\sqrt{1-x^4}+2E(sin^{(-1)}(x)|-1)+2\frac{\sqrt{x²}}{x}E(\frac{cos^{(-1)}(x²)}2|2)}3$+C
when I first tested this formula (yes, I already had the proof written out, literally the other day I had done this by coincidence haha), my results were that the answer isn't $\frac{\Gamma(\frac{1}4)²}{6\sqrt{2π}}$, however, after the issue with the "sign of the real number of x", I had found that your answer is correct, but sadly my formula doesn't have any conclusive evidence to prove it.  But I'm not going to delete this because #1 it's REALLY hard writing out all the formulas in who_cares_what_language++ on a mobile phone, #2 because in between checking that all my formulas are written correctly by submitting and resubmitting, I had noticed some comments on my answer, here's my response to them: I know your trying to be as helpful as possible, and I'm glad you're doing that, but everyone on this website needs to stop being so passive aggressive towards one another.  Everyone on here is competing to be "more right" than one another, but that's not how math works. You're either right or wrong, and if you're wrong, that doesn't mean you're dumb, it means your trying new ideas, and sometimes your ideas don't work, or sometimes they can't be verified.  Nevertheless, making mistakes isn't a sign of ignorance, it's a sign of innovation, and while I'm sorry my answer wasn't of any help to you, I know that someone on here will probably use my work to help them think of a different approach.  So I hope you'll take that evidence to heart, because it's not a good sign when even on a website that's strictly for intellectual purposes, were still not immune to the petty "I'm right, you're wrong, hahaha" attitude of the internet.  Yes, it IS important to correct others mistakes when they're not quite right, but you have to be respectful, stop questioning eachother's intelligence.  And most importantly, GOSH DARN is it hard to write equations on here!!!
-Math Machine
