Proving that $\iint\limits_{x^2+y^2\leq 1} e^x \cos y dxdy=\pi$

Write

\begin{align*} 2 \int_{-1}^{1} e^x \sin \sqrt{1-x^2} \, dx &= 2 \Im \int_{-1}^{1} e^{x+i\sqrt{1-x^2}} \, dx \\ &= 2 \Im \int_{0}^{\pi} e^{e^{i\theta}}\sin\theta \, d\theta \qquad (x = \cos\theta) \\ &= \Im \int_{0}^{\pi} i e^{e^{i\theta}} (e^{-i\theta} - e^{i\theta}) \, d\theta. \end{align*}

Now expanding the double exponential term using the Taylor series, we can write

\begin{align*} 2 \int_{-1}^{1} e^x \sin \sqrt{1-x^2} \, dx &= \sum_{n=0}^{\infty} \frac{1}{n!} \Im \int_{0}^{\pi} i (e^{(n-1)i\theta} - e^{(n+1)i\theta}) \, d\theta. \end{align*}

But notice that we have the following sifting property:

$$ \Im \int_{0}^{\pi} i e^{ik\theta} \, d\theta = \begin{cases} \pi, & k = 0 \\ 0, & k \neq 0 \end{cases}. $$

Therefore we get

$$ 2 \int_{-1}^{1} e^x \sin \sqrt{1-x^2} \, dx = \frac{1}{0!}(0 - 0) + \frac{1}{1!}({\color{red}\pi} - 0) + \frac{1}{2!}(0 - 0) + \cdots = \pi. $$

It is not that crazy. It is equivalent to $$ I = 2\cdot\text{Im}\int_{0}^{1}\exp\left(x+i\sqrt{1-x^2}\right)+\exp\left(-x+i\sqrt{1-x^2}\right)\,dx $$ or to $$ I = 2\cdot\text{Im}\int_{0}^{\pi/2}\left[\exp(e^{i\theta})+\exp(-e^{-i\theta})\right]\sin(\theta)\,d\theta$$ or, by exploiting $e^z=\sum_{n\geq 0}\frac{z^n}{n!}$, to $$ I = 2\cdot\sum_{n\geq 0}\frac{1}{n!}\int_{0}^{\pi/2}\left[\sin(n\theta)-(-1)^n \sin(n\theta)\right]\sin(\theta)d\theta $$ that simplifies to: $$ I = 4\cdot\sum_{m\geq 0}\frac{1}{(2m+1)!}\int_{0}^{\pi/2}\sin((2m+1)\theta)\sin(\theta)\,d\theta $$ where only the contribute given by $m=0$ is non-zero. It follows that: $$ I = 4\int_{0}^{\pi/2}\sin^2(\theta)\,d\theta = \color{red}{\pi}.$$