<s>[INST] 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 $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?
 [/INST] A slight variant of lab bhattacharjee's method provides a simpler solution:
Let
$$ I(a) = \int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx = 2\int_{0}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx. $$
Then by a simple application of multivariable calculus,
\begin{align*}
I'(a)
&= 2 \left. \frac{\sqrt{a^2-x^2}}{1+x^2} \right|_{x=a} + 2 \int_{0}^{a} \frac{d}{da} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx \\
&= 2 \int_{0}^{a} \frac{a}{(1+x^2)\sqrt{a^2-x^2}} \, dx.
\end{align*}
Then with the change of variable $x = a \sin\theta$, we have
\begin{align*}
I'(a)
&= 2 \int_{0}^{\frac{\pi}{2}} \frac{a}{1+a^2\sin^2\theta} \, d\theta \\
&= 2 \int_{0}^{\frac{\pi}{2}} \frac{a \sec^2\theta}{1+(a^2+1)\tan^2\theta} \, d\theta \\
&= 2 \int_{0}^{\infty} \frac{a}{1+(a^2+1)t^2} \, dt \qquad (t = \tan\theta) \\
&= \frac{\pi a}{\sqrt{a^2+1}}.
\end{align*}
Thus by integrating, we must have
$$ I(a) = \pi\sqrt{a^2+1} + C $$
for some constant $C$. But
$$ I(0+) = \lim_{a\to0} \int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx  = \lim_{a\to0} \int_{-1}^{1} \frac{\sqrt{1-x^2}}{(1/a)^2+x^2} \, dx = 0 $$
and we must have $C = -\pi$. This proves the identity.
</s>