Prove that the integral of $e^{k\cos t}\cos(k\sin t)$ from $0$ to $2\pi$ equals $2\pi$

Question

Prove that $$I=\int_0^{2\pi}e^{k\cos t}\cos(k\sin t)\,dx=2\pi$$ and prove that $$J=\int_0^{2\pi}e^{k\cos t}\sin(k\sin t)\,dx=0$$ With the help of the following integral: $$H=\int_{|z|=1}\frac{e^{kz}}{z}\,dx$$ I proved that the : $$H=\int_{|z|=1}\frac{e^{kz}}{z}\,dx=2i\pi$$ and I proved that $$I+iJ=\frac{H}{i}=2\pi$$but what should I do next to prove that$I=2\pi$ or to prove that$J=0$? Is it by using $Re(z)=\cos t$?

Answer 1

I proved that the : $$H=\int_{|z|=1}\frac{e^{kz}}{z}\,dx=2i\pi$$ and I proved that $$I+iJ=\frac{H}{i}=2\pi$$but what should I do next to prove that$I=2\pi$ or to prove that$J=0$? Since $ t $ is Real Number, both of the integrals $I$ and $J$ are Real. Which means we can compare both sides of the equation $$I+iJ=2\pi$$ , $$I+iJ=2\pi+0i$$ which leads to $$I=2\pi$$ $$J=0$$. End of solution.