Determine area bounded by $(x+y)^4 = ax^2y$

Question

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?

Answer 1

\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}