I have the following equation: $$\left(x+y\right)^5 =x^2y^2.$$ I need to find the area bounded by this equation, I used: $x=r\cos(\theta)$, $y=r\sin(\theta)$. I then got $$r=\frac{\sin^2(\theta)\cos^2(\theta)}{(\cos(\theta)+\sin(\theta)^5}$$ So the graph looks like this:
I wrote my integral as $$\int_{0}^{\pi/2}d\phi\int_{0}^{r}rdr$$ which then equals to: $$\frac{1}{2}\int_{0}^{\pi/2}\frac{\sin^4(\phi)\cos^4(\phi)}{(\cos(\phi)+\sin(\phi))^{10}}\,d\phi.$$ I'm not sure if I am correct to this point, and I'm not able to find a substitution to solve this integral.
The integral looks fine to me. To evaluate it, divide the numerator and the denominator by $\cos^{10}\phi$ and then use the substitution $u=\tan\phi, \mathrm du=\frac{\mathrm d\phi}{\cos^2\phi}$ to obtain $$\frac12\int_0^{\frac{\pi}2}\frac{\sin^4\phi\cos^4\phi}{(\cos\phi+\sin\phi)^{10}}\mathrm d\phi=\frac12\int_0^{\frac{\pi}2}\frac{\tan^4\phi}{(1+\tan\phi)^{10}}\frac{\mathrm d\phi}{\cos^2\phi}=\frac12\int_0^\infty\frac{u^4}{(1+u)^{10}}\mathrm d u.$$ Now use the substitution $t=1+u,\mathrm dt=\mathrm du$ and expand the numerator.