Prove: $$A=\int_{0}^\pi{e^{a\cos(x)}\cos(a\sin(x))}dx=\pi$$
(I think) that a suggestion was made to calculate and later use it: $B=\int\frac{{e^{az}}}{z}$, over the path gamma $\gamma=e^{it}, t\in[-\pi,\pi]$, i think this integral is $2i\pi$ for all a, because of Cauchy theorem (correct me please if I am wrong) , then I tried to make it look similar to the second integral: $$B=\int_{-\pi}^\pi e^{a\cos(x)}dx+\int_{-\pi}^\pi e^{i\sin(x)}dx,$$ but do not know how to proceed. Any help appreciated
From $$ 2\,\pi\,i=\int_\gamma\frac{e^{iaz}}{z}\,dz=\int_0^{2\pi}\frac{e^{ae^{it}}}{e^{it}}\,i\,e^{it}\,dt $$ we get $$ \int_0^{2\pi}e^{ae^{it}}\,dt=2\,\pi. $$ Now $$ e^{ae^{it}}=e^{a\cos t+ia\sin t}=e^{a\cos t}\bigl(\cos(a\,\sin t)+i\cos(a\,\sin t)\bigr). $$ Since $2\pi$ is real, we get $$ \int_0^{2\pi}e^{a\cos t}\cos(a\,\sin t)\,dt=2\,\pi. $$