Showing that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left (\sqrt{a^2+1}-1\right)$. How can I show that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?

You can also convert it into an integral over the upper half-disk $D_a$ of radius $a$:
$$\int_{D_a} {1 \over 1 + x^2}\,dy\,dx$$
(Do the $y$ integral first to convert it into the original form.) Since the integrand is even, it is twice the integral over the portion where $x > 0$.
Changing to polar coordinates, this becomes
$$2\int_0^a \int_0^{\pi \over 2} {r \over 1 + r^2\cos^2(\theta)}\,d\theta\,dr$$
$$= 2\int_0^a r \int_0^{\pi\over 2} {\sec^2(\theta) \over \sec^2(\theta) + r^2}\,d\theta\,dr$$
$$= 2\int_0^a r \int_0^{\pi\over 2} {\sec^2(\theta) \over \tan^2(\theta) + (1 + r^2)}\,d\theta\,dr$$
Doing a $u$ substitution this becomes
$$= 2\int_0^a r \int_0^{\infty} {1 \over u^2 + (1 + r^2)}\,du\,dr$$
$$= 2\pi\int_0^a r {1 \over \sqrt{1 + r^2}}\,dr$$
$$=\pi \sqrt{1 + r^2}\,\,\,\bigg|_{r = 0}^{r = a}$$
$$=\pi\sqrt{1 + a^2} - \pi$$
