# Definite Integral $\int_{-\pi}^{\pi}\frac{e^{\sin(x)+\cos(x)}\cos(\sin(x))}{e^{\sin(x)}+e^x}dx$

How can I find the value of this following definite integral?

$$\int_{-\pi}^{\pi}\frac{e^{\sin(x)+\cos(x)}\cos(\sin(x))}{e^{\sin(x)}+e^x}dx$$

Mathematica says it's value is $\pi$, but I don't know how obtain it?

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Let $x\mapsto -x$ –  L. F. Nov 23 '13 at 13:21
@L.F. What do you mean, please ? I am really mpade curious by this post. I always thought that teachers imagination has no limit. –  Claude Leibovici Nov 23 '13 at 13:28

$$x\mapsto -x:$$ $$\int_{-\pi}^{\pi} \frac{e^{\sin x+\cos x}\cos \sin x}{e^{\sin x}+e^{x}}\,dx=\int_{-\pi}^{\pi} \frac{e^{ x+\cos x}\cos \sin x}{e^{\sin x}+e^{x}}\,dx$$

$$\frac{1}{2}\int_{-\pi}^{\pi}e^{\cos x}\cos \sin x\,dx=\frac{1}{2}\int_0^{2\pi}e^{-\cos x}\cos \sin x\,dx=\pi$$

The last line can be done as follows:

$$\int_0^{2\pi} e^{-\cos x}\cos(- \sin x)\,dx=\Re\left(\int_0^{\pi} e^{-e^{ix}dx}\right)$$

$$f(\lambda)=\int_0^{2\pi} e^{\lambda e^{ix}}dx$$

$$f'(\lambda)=\frac{1}{i\lambda}\int_0^{2\pi} i\lambda e^{i x}e^{\lambda e^{ix}}dx =\frac{1}{i\lambda}\bigg[e^{\lambda e^{ix}}\bigg]_0^{2\pi}=0$$

Hence:

$$f(\lambda)=\mathcal{C}$$

Taking $\lambda =0$ we have $\mathcal{C}=2\pi$ and the result follows.

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Let $u=-x$. Denote your integral as $I$. Then

$$I=\int_{-\pi}^{\pi}\frac{\exp{(-\sin u+\cos u)}\cos(\sin u)}{\exp{(-u)}+\exp{(-\sin u)}}du=\int_{-\pi}^{\pi}\frac{\exp{(u+\cos u)}\cos(\sin u)}{\exp{(u)}+\exp{(\sin u)}}du$$

Then $$2I=\int_{-\pi}^{\pi}\frac{(\exp{(\sin u)+\exp(u))\exp(\cos u)}\cos(\sin u)}{\exp{(u)}+\exp{(\sin u)}}du=\int_{-\pi}^{\pi}\exp{(\cos u)}\cos(\sin(u))du$$

The following argument is an application of complex residue:

$$\int_{C}\frac{\exp z}{z}dz=2\pi i,$$

where C is the circle $\vert z\vert=1$.

Let $z=\exp{i\theta} (-\pi<\theta<\pi),$ then the integral is transformed into

$$\int_{-\pi}^{\pi}\exp{(\exp(i\theta))}d\theta=2\pi$$

Recall that $$\exp{(\exp(i\theta))}=\exp{(\cos\theta)}(\cos(\sin\theta)+i\sin(\sin\theta))$$

then the result follows.

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