# Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$

Integrate

$$I=\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$$

Substitution $x=\sqrt{\tan(u)}\rightarrow dx={\sec^2(u)\over 2\sqrt{\tan(u)}}du$

$x=1\rightarrow u={\pi\over 4}$

$x=0\rightarrow u=0$

$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-\tan^2(u)}\over 1+\tan^2(u)}\cdot {\sec^2(u)\over \sqrt{\tan(u)}}du$$

$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-\tan^2(u)\over \tan(u)}}du$$

$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{\cot(u)-\tan(u)}}du$$ Recall

$$\cot(u)-\tan(u)={\cos^2(u)-\sin^2(u)\over \sin(u)\cos(u)}=2\cot(2u)$$

Substitute back into I

$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{2\cot(2u)}}du$$

$$I={\sqrt2\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{\cot(2u)}}du$$

Well I know that $$\int{\cos(2u)\over\sin(2u)}du={1\over 2}\ln(\sin(2u))+C$$ but

$$\int\sqrt{{\cos(2u)\over\sin(2u)}}du$$ I have not idea, so can anyone please give a hand? Thank.

• Well here is a definite integral and other is an indefinite integral. That is the only difference I can see from it otherwise it is a duplicate. – gymbvghjkgkjkhgfkl Jun 4 '16 at 22:20
• There is also a difference in the signs in front of $x^4$. – Hans Lundmark Jun 5 '16 at 6:15
• You are right @HansLundmark how did I missed that! – gymbvghjkgkjkhgfkl Jun 5 '16 at 8:16
• I have withdrawn my duplication comment, but I think it should be mentioned that I copied your integrand to the M.SE search engine and it "matched" to posts with $\ \int \ \frac{\sqrt{1+x^4}}{1-x^4} \ dx \$ . I should have checked back more carefully against your actual integral, but it also indicates that searches are not necessarily reliable here... – colormegone Jun 5 '16 at 22:16

You were so close! \begin{align}\int_0^{\frac{\pi}4}\sqrt{\cot2u}\,du&=\frac12\int_0^{\frac{\pi}2}\sqrt{\cot v}\,dv\tag{1}\\ &=\frac14\cdot2\int_0^{\frac{\pi}2}\cos^{1/2}v\sin^{-1/2}v\,dv\tag{2}\\ &=\frac14\text{B}\left(\frac34,\frac14\right)\tag{3}\\ &=\frac14\frac{\Gamma\left(\frac34\right)\Gamma\left(\frac14\right)}{\Gamma(1)}\tag{4}\\ &=\frac14\frac{\pi}{\sin\frac{\pi}4}\tag{5}\\ &=\frac{\pi}{2\sqrt2}\tag{6}\end{align} EDIT: Some annotations:
$(1)$ Let $2u=v$
$(2)$ $\cot\theta=\frac{\cos\theta}{\sin\theta}$
$(3)$ Recognizing the trigonometric form of the Beta function
$(4)$ Using the relationship between the Beta and Gamma functions
$(5)$ The reflection formula for the Gamma function. We might equally have used the duplication formula $$\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)=\sqrt{\pi}2^{1-2\left(\frac14\right)}\Gamma\left(2\left(\frac14\right)\right)=\sqrt{\pi}\sqrt2\sqrt{\pi}=\pi\sqrt2$$ $(6)$ Of course $\sin\frac{\pi}4=\frac1{\sqrt2}$. At this point we're home free and we can substitute the value of the integral into where you were stuck and arrive at $$\int_0^1\frac{\sqrt{1-x^4}}{1+x^4}dx=\frac{\sqrt2}2\frac{\pi}{2\sqrt2}=\frac{\pi}4$$

• Why the downvote? I showed how to get from where he was stuck to the solution. Isn't that acceptable any more? – user5713492 Jun 5 '16 at 0:56
• Yeah why the vote down! This is the point I am looking for. (+1) @user5713492. – gymbvghjkgkjkhgfkl Jun 5 '16 at 3:26
• How did you get from 1st integral into the 2nd integral? – gymbvghjkgkjkhgfkl Jun 5 '16 at 3:32
• I added some annotations in case my proof wasn't completely clear. Sometimes people try to disrupt the forum by downvoting randomly selected posts. – user5713492 Jun 5 '16 at 5:32

Let $y = x^2$ and $z = \sqrt{y^{-1} - y}$, we have \begin{align} \int_0^1 \frac{\sqrt{1-x^4}}{1+x^4}dx = & \int_0^1 \frac{x\sqrt{x^{-2} - x^2}}{x^2(x^{-2}+x^2)}dx = \int_0^1 \frac{\sqrt{x^{-2} - x^2}}{(x^{-2}+x^2)}\frac{dx}{x}\\ = & \frac12 \int_0^1 \frac{\sqrt{y^{-1}-y}}{y^{-1}+y}\frac{dy}{y} = -\frac12 \int_{y=0}^1 \frac{\sqrt{y^{-1}-y}}{y^{-1}+y}\frac{d(y^{-1} - y)}{y^{-1}+y}\\ = & \int_0^\infty \frac{z^2 dz}{z^4+4} = \int_0^\infty \frac{z^2 dz}{(z^2+2)^2 - (2z)^2} = \int_0^\infty \frac{z^2 dz}{(z^2-2z+2)(z^2+2z+2)}\\ = & \frac14 \int_0^\infty\left(\frac{z}{z^2-2z+2} - \frac{z}{z^2+2z+2}\right)dz\\ = & \frac14 \int_0^\infty \left[\frac12\log\left(\frac{z^2-2z+2}{z^2+2z+2}\right)' + \left(\frac{1}{z^2-2z+2} + \frac{1}{z^2+2z+2}\right)\right] dz \end{align} The first piece contributes $\displaystyle\;\frac18 \left[\log\left(\frac{z^2-2z+2}{z^2+2z+2}\right)\right]_0^\infty\;$ which clearly vanishes.

For the second piece, substitute $z$ by $-z$ in its second term, we get:

$$\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4}dx = \frac14 \int_{-\infty}^\infty\frac{dz}{z^2-2z+2} = \frac14 \int_{-\infty}^\infty\frac{dz}{(z-1)^2+1} = \frac{\pi}{4}$$

On the path of Achille Hui,

$\displaystyle J=\int_0^{\infty} \dfrac{x^2}{x^4+4}dx$

Perform the change of variable $y=\dfrac{x}{\sqrt{2}}$,

$\displaystyle J=\dfrac{\sqrt{2}}{2}\int_0^{\infty} \dfrac{x^2}{x^4+1}dx$

1)

Perform the change of variable $y=x^4$,

\begin{align} J&=\dfrac{\sqrt{2}}{8}\int_0^{\infty} \dfrac{x^{-\tfrac{1}{4}}}{1+x}dx\\ &=\dfrac{\sqrt{2}}{8}\text{B}\left(\dfrac{3}{4},\dfrac{1}{4}\right)\\ &=\dfrac{\sqrt{2}}{8}\dfrac{\Gamma\left(\dfrac{3}{4}\right)\Gamma\left(\dfrac{1}{4}\right)}{\Gamma\left(1\right)}\\ &=\dfrac{\sqrt{2}}{8}\dfrac{\pi}{\sin\left(\tfrac{\pi}{4}\right)}\\ &=\dfrac{\sqrt{2}}{8}\dfrac{\pi}{\tfrac{\sqrt{2}}{2}}\\ &=\boxed{\dfrac{\pi}{4}} \end{align}

2) alternatively

$\displaystyle J=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty} \dfrac{x^2}{1+x^4}dx$

Perform the change of variable $y=\dfrac{1}{x}$,

$\displaystyle J=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty} \dfrac{1}{1+x^4}dx$

therefore,

$\displaystyle 2J=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty}\dfrac{1+\tfrac{1}{x^2}}{x^2+\tfrac{1}{x^2}}dx$

Observe that,

$x^2 +\dfrac{1}{x^2}=\left(x-\dfrac{1}{x}\right)^2+2$

Perform the change of variable $y=x-\dfrac{1}{x}$,

\begin{align} 2J&=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty} \dfrac{1}{x^2+2}dx\\ &=\dfrac{\sqrt{2}}{4}\left[\frac{\arctan\left(\frac{x}{\sqrt{2}}\right)}{\sqrt{2}}\right]_{-\infty}^{+\infty}\\ &=\dfrac{\pi}{2} \end{align}

therefore,

$\boxed{J=\dfrac{\pi}{4}}$