Determine area bounded by $(x+y)^4 = ax^2y$ Determine the area of region $U$ bounded by the graphic of the curve:
$$(x+y)^4 = ax^2y,\ a > 0 \quad \text{(loop in the first quadrant)}$$
I used polar coordinates and I arrived to:
$$\iint r\,dr\,d\theta = \frac{a^2}{2}\int_0^{\frac{\pi}{2}}
\frac{\cos^4 \theta \,\sin^2 \theta} {(\cos \theta + \sin \theta)^8}
\,d\theta$$
but I don't know how to compute this integral. Any hint for it?

\begin{align}
&\int_0^{\pi/2}
\frac{\cos^4 \theta \,\sin^2 \theta} {(\cos \theta + \sin \theta)^8}
\,d\theta\\
=& \int_0^{\pi/2}
\frac{\tan^2 \theta \,\sec^2 \theta} {(1+ \tan \theta)^8}
\,d\theta
\overset{x=\tan\theta}=\int_0^{\infty}
\frac{x^2}{(1+x)^8}dx\\
=& \int_0^{\infty}
\left( \frac{1}{(1+x)^6}-\frac{2}{(1+x)^7}+ \frac{1}{(1+x)^8}\right) dx=\frac{1}{105}
\end{align}
