Double Integral $(x^2+y^2)$ over region $x^4+y^4 \le 1$ I have calculus exercise to calculate double integral of function $f(x)=x^2+y^2$ over the area enclosed inside curve $x^4+y^4=1$. I have tried with polar coordinates:
$$
\iint_D f(\phi,r)r \,d \phi\,dr = 4\int_{0}^{\pi/2} \int_{0}^{[1/(\cos^4\phi+\sin^4\phi)]^{1/4}} r^3 d\phi dr
$$
Although this seems to be the right way (I get the right result with Wolfram Mathematica), it leads to the integral
$$
\int_{0}^{\pi/2}\frac{dx}{\cos^4x+\sin^4x}
$$
which I don't know how to easily execute.
I was wondering if there is any trick to use diferent new coordinate systems or integration by substitution? Is there any general trick to integrate a function ower the area of this curve, because it appears quite often in exercises?
 A: 
I thought I would present an approach that appeals to the Beta and Gamma functions to facilitate the valuation.   To that end, we  now proceed.


From symmetry considerations, we can write
$$\begin{align}
\iint_{x^4+y^4\le 1}(x^2+y^2)\,dx\,dy&=8\int_0^1 \int_0^{(1-y^4)^{1/4}}x^2\,dx\,dy\\\\
&=\frac83\int_0^1 (1-y^4)^{3/4}\,dy\\\\
&=\frac23\int_0^1 (1-t)^{3/4}t^{-3/4}\,dt \tag1\\\\
&=\frac23 B\left(\frac74,\frac14\right)\tag2\\\\
&=\frac23 \Gamma(7/4)\Gamma(1/4)\tag3\\\\
&=\frac12\Gamma(3/4)\Gamma(1-3/4)\tag4\\\\
&=\frac12\frac\pi{\sin(3\pi/4)}\tag5\\\\
&=\frac\pi{\sqrt 2}
\end{align}$$

NOTES:
In arriving at $(1)$, we made the substitution $y=t^{1/4}$.
In arriving at $(2)$ we recognized $(1)$ as the "standard" integral representation of the Beta function.
In going from $(2)$ to $(3)$ we used the relationship $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ between the Beta and Gamma functions.
In going from $(3)$ to $(4)$ we made use of the functional equation $\Gamma(1+x)=x\Gamma(x)$.
In going from $(4)$ to $(5)$ we used Euler's Reflection formula $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$
A: Note $\cos^4x+\sin^4x=1-2\cos^2x\sin^2x= \frac12(1+\cos^22x)$
\begin{align}
&\int_{0}^{\pi/2}\frac{1}{\cos^4x+\sin^4x}dx\\
= &\int_{0}^{\pi/2}\frac{2dx}{1+\cos^22x}
\overset{2x\to x}= \int_{0}^{\pi}\frac{dx}{1+\cos^2x}
=  2\int_{0}^{\pi/2}\frac{d(\tan x)}{2+\tan^2x}=\frac\pi{\sqrt2}
\end{align}
A: Welcome to MSE. We have
$$
\int_{0}^{\pi/2}\frac{1}{\cos^4(x)+\sin^4(x)}\,dx = \int_{0}^{\pi/2}\frac{\sec^4(x)}{1+\tan^4(x)}\,dx
$$Put $z=\tan(x)$, $dz=\sec^2(x)\,dx$. Then $\sec^2(x)=z^2+1$:
$$
\Rightarrow \int _{0}^{\infty} \frac{1+z^2}{1+z^4}\,dz
$$Now use partial fractions (I'll skip the steps):
$$
=\frac{1}{2}\int _0^{\infty} \frac{1}{z^2+\sqrt{2}z+1}+\frac{1}{z^2-\sqrt{2}z+1}\,dz
$$Complete the square and integrate using arctangent; you can do the rest.
A: First, symmetry of $x^2+y^2$ tells us
$$\iint_{x^4+y^4\le1}x^2+y^2\,\mathrm dx\,\mathrm dy=4\iint_Dx^2+y^2\,\mathrm dx\,\mathrm dy$$
where $D$ is the part of the region $x^4+y^4\le1$ in the first quadrant.
Use a different change of coordinates:
$$\begin{cases}x=\sqrt{r\cos\theta}\\y=\sqrt{r\sin\theta}\end{cases}$$
which ensures $x^2+y^2=r(\cos\theta+\sin\theta)$. Then we get $D$ with $0\le r\le1$ and $0\le\theta\le\frac\pi2$.
The Jacobian determinant for this transformation is
$$\det\begin{bmatrix}(\sqrt{r\cos\theta})_r&(\sqrt{r\cos\theta})_\theta\\
(\sqrt{r\sin\theta})_r&(\sqrt{r\sin\theta})_\theta\end{bmatrix}=\frac14\csc\theta\sqrt{\tan\theta}$$
and the integral is
$$\begin{align*}
4\iint_Dx^2+y^2\,\mathrm dx\,\mathrm dy&=\int_0^{\pi/2}\int_0^1r(\cos\theta+\sin\theta)\csc\theta\sqrt{\tan\theta}\,\mathrm dr\,\mathrm d\theta\\[1ex]
&=\frac12\int_0^{\pi/2}(\cot\theta+1)\sqrt{\tan\theta}\,\mathrm d\theta
\end{align*}$$
Substitute $u=\sqrt{\tan\theta}$, or $u^2=\tan\theta$. Notice that $1+u^4=\sec^2\theta$. Compute the differential:
$$2u\,\mathrm du=\sec^2\theta\,\mathrm d\theta\implies\mathrm d\theta=\frac{2u}{1+u^4}\,\mathrm du$$
Then we arrive at the same integral as user Integrand:
$$\begin{align*}
\frac12\int_0^{\pi/2}(\cot\theta+1)\sqrt{\tan\theta}\,\mathrm d\theta&=\frac12\int_0^\infty\left(\frac1{u^2}+1\right)u\frac{2u}{1+u^4}\,\mathrm du\\[1ex]
&=\int_0^\infty\frac{1+u^2}{1+u^4}\,\mathrm du
\end{align*}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\iint_{\large\mathbb{R}^{2}}\bracks{x^{4} + y^{4} < 1}
\pars{x^{2} + y^{2}}\dd x\,\dd y}
\\[5mm] = &\
4\int_{0}^{\infty}\int_{0}^{\infty}\bracks{x^{4} + y^{4} < 1}
\pars{x^{2} + y^{2}}\dd x\,\dd y
\\[5mm] \stackrel{\substack{x^{2}\ \mapsto\ x \\ y^{2}\ \mapsto\ y}}{=}\,\,\, &
\int_{0}^{\infty}\int_{0}^{\infty}\bracks{x^{2} + y^{2} < 1}
\pars{x + y}{\dd x \over \root{x}}\,{\dd y \over \root{y}}
\\[5mm] = &\
\int_{0}^{\pi/2}\int_{0}^{1}
{\cos\pars{\phi} + \sin\pars{\phi} \over
\root{\sin\pars{\phi}\cos\pars{\phi}}}r\,\dd r\,\dd\phi
\\[5mm] = &\
{1 \over 2}\root{2}\int_{0}^{\pi/2}
\root{1 + \sin\pars{2\phi} \over \sin\pars{2\phi}}\dd\phi =
{1 \over 4}\root{2}\int_{0}^{\pi}
\root{1 + \sin\pars{\phi} \over \sin\pars{\phi}}\dd\phi
\\[5mm] = &\
{1 \over 4}\root{2}\int_{-\pi/2}^{\pi/2}
\root{1 + \cos\pars{\phi} \over \cos\pars{\phi}}\dd\phi =
{1 \over 2}\root{2}\int_{0}^{\pi/2}
\root{1 + \cos\pars{\phi} \over \cos\pars{\phi}}\dd\phi
\\[5mm] = &\
{1 \over 2}\root{2}\int_{0}^{\pi/2}
{\sin\pars{\phi} \over \root{\cos\pars{\phi} - \cos^{2}\pars{\phi}}}\dd\phi
\\[5mm] = &\
{1 \over 2}\root{2}\
\underbrace{\int_{0}^{1}{\dd\xi \over \root{\xi - \xi^{2}}}}
_{\ds{\pi}}\ = \bbx{{\root{2} \over 2}\,\pi}
\approx 2.2214 \\ &
\end{align}
