Can this integral be evaluated ? If Yes, then How?: $\int_0 ^{\pi/2} e^{\cos(\tan x)}\cos(\sin(\tan x))\space dx$ $$\int_0 ^{\pi/2} e^{\cos(\tan x)}\cos(\sin(\tan x))\space dx$$
Please use simple techniques and properties (if possible).
 A: Here is a simpler form of the integral: $$\frac12 \int_0^\frac\pi2 e^{e^{i\tan(x)}}dx+ \frac12 \int_0^\frac\pi2 e^{e^{-i\tan(x)}}dx=\text{Re}\int_0^\frac\pi2e^{e^{i\tan(x)}}dx$$
Now expand by using a series:
$$\text{Re}\int_0^\frac\pi2e^{e^{i\tan(x)}}dx = \text{Re}\int_0^\frac\pi2\left(1-\sum_{n=1}^\infty\frac{e^{in\tan(x)}}{n!}\right)dx$$
Using the Ei function, the substitution $x=\tan^{-1}(t)$, partial fractions, and integration to get:
$$\text{Re}\int_0^\frac\pi2\left(1-\sum_{n=1}^\infty\frac{e^{in\tan(x)}}{n!}\right)dx = \text{Re}\left(\frac\pi2-\frac i2\sum_{n=1}^\infty\frac{e^n\text{Ei}((i\tan(x)-1)n)-e^{-n}\text{Ei}((i\tan(x)+1)n)}{n!}\right)\bigg|_0^\frac\pi2$$
taking limits for the evaluations:
$$\text{Re}\left(\frac\pi2-\frac i2\sum_{n=1}^\infty\frac{e^n\text{Ei}((i\tan(x)-1)n)-e^{-n}\text{Ei}((i\tan(x)+1)n)}{n!}\right) = \text{Re}\left(\frac\pi2-\sum_{n=1}^\infty\frac{\pi\sinh(n)-\frac12\left(e^n(i\text{Ei}(-n)+\pi\right)-ie^{-n}\text{Ei}(n))}{n!}\right)= \frac\pi2- \frac\pi2\left(e^e-\sqrt[e]e\right)-\sum_{n=1}^\infty \frac{\pi e^n}{2n!}$$
which is checked here. Now we verify @J.G.’s solution:
$$\boxed{\int_0^\frac\pi2e^{\cos(\tan(x))}\cos(\sin(\tan(x)))dx=\frac{\pi\sqrt[e]e}2}$$
However, the bonus $\int_0^\frac\pi2e^{\cos(\tan(x))}\sin(\sin(\tan(x)))dx$ has no closed form seen in the bolded link.
