<s>[INST] Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ . Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ . 
Answer:
The graph is above : 
         
Since the region is symmetrical , 
$  Area =4  \times \int_{0}^{2} \int_{0}^{\sqrt{2+\sqrt{4x^2-x^4+4}}} dxdy $
Am I right ? Is there any help ? 
 [/INST] Convert to polar form $r^2=\dfrac{16}{\cos 4 t+3}$
The area is $$A=\frac{1}{2} \int_0^{2 \pi } \frac{16}{\cos 4 t+3} \, dt=32 \int_0^{\frac{\pi }{2}} \frac{1}{\cos 4 t+3} \, dt$$
$\cos 4t=\cos^2 2t-\sin^2 2t=2\cos^2 2t -1$
so we have to integrate
$\int \dfrac{dt}{2(\cos^2 2 t+1)}=\dfrac{1}{2}\int \dfrac{dt}{1+\cos^2 2t}$
Remember the identity $\cos 2t =\dfrac{1-\tan^2 t}{1+\tan^2 t}$
with the substitution $\tan t=u;\;t=\tan^{-1}u;\;dt=\dfrac{du}{1+u^2};\;\cos 2t = \dfrac{1-u^2}{1+u^2}$ 
Limits of integration become $t=0\to u=0;\;t=\pi/2\to u=\infty$
$$\dfrac{1}{2}\int \dfrac{dt}{1+\cos^2 2t}=\dfrac{1}{2}\int \dfrac{\dfrac{du}{1+u^2}}{1+\left(\dfrac{1-u^2}{1+u^2}\right)^2}$$
Which simplifies to
$$\dfrac{1}{2}\int \frac{u^2+1}{ u^4+1} \, du=\frac{\tan ^{-1}\left(\sqrt{2} u+1\right)-\tan ^{-1}\left(1-\sqrt{2} u\right)}{4 \sqrt{2}}$$
Area is
$$A=32\left[\frac{\tan ^{-1}\left(\sqrt{2} u+1\right)-\tan ^{-1}\left(1-\sqrt{2} u\right)}{4 \sqrt{2}}\right]_0^{\infty}=4 \sqrt{2} \pi$$
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