How could I solve the following problem to compute: $$\int_{R}x^2+y^2\,dx\,dy$$ over region R bounded by the polar curve $r=1+\cos{\theta}$ from $\theta \in [0,pi]$ and the x-axis?
I have attempted to apply a polar change of variable, but have only managed to reduce it to: $$\int_{0}^{\pi}4\cos^{8}{\frac{\theta}{2}}\, d\theta$$ which is clearly not simple enough.
Note that we have
$$\begin{align} \iint_R (x^2+y^2)\,dx\,dy&=\int_0^\pi \int_{0}^{1+\cos(\theta)}r^3 \,dr\,d\theta\\\\ &=\frac14\int_0^\pi (1+\cos(\theta))^4\,d\theta\\\\ &=4\int_0^\pi \cos^8(\theta/2)\,d\theta\\\\ &=8\int_0^{\pi/2} \cos^8(\theta)\,d\theta \end{align}$$
Using the reduction formula for the integral $\int_0^{\pi/2} \cos^n(\theta)\,d\theta=\frac{n-1}{n}\int_0^{\pi/2}\cos^{n-2}(\theta)\,d\theta$, we find that
$$8\left(\int_0^{\pi/2} \cos^8(\theta)\,d\theta\right)=8\left(\frac78 \cdot \frac56\cdot\frac34\cdot\frac12\cdot \frac{\pi}{2}\right)=\frac{35\pi}{32}$$