# Basic Integration question

Prove $\exists \xi \in [0,\pi]$ such that $$\int_0^\pi e^{-x}\cos(x)\,dx = \sin(\xi)$$

Well, I was thinking on using the Generalized MVT, but it doesnt seem I obtain the right answer. How do you guys would tackle this problem?

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$$\left|\int_0^\pi\mathrm e^{-x}\cos(x)\,\mathrm dx\right|\lt\int_0^\pi\mathrm e^{-x}\,\mathrm dx=\left[-\mathrm e^{-x}\right]_0^\pi=1-\mathrm e^{-\pi}\lt1\;.$$
Actually if $\xi$ is supposed to be in $[0,\pi]$, you also want it to be $\ge 0$. For that, apply the change of variables $u =\pi - x$ to obtain $$\int_{\pi/2}^\pi e^{-x} \cos(x)\ dx = -\int_0^{\pi/2} e^{x-\pi} \cos(x)\ dx$$ so $$\int_0^\pi e^{-x} \cos(x)\ dx = \int_0^{\pi/2} (e^{-x} - e^{x-\pi}) \cos(x)\ dx \ge 0$$ since $e^{-x} - e^{x-\pi}$ for $0 \le x \le \pi/2$. – Robert Israel Oct 20 '12 at 0:02
Note that $$\int_0^\pi {{e^{ - x}}\cos (x)dx} = \frac{{{e^{ - \pi }}(1 + {e^\pi })}}{2}$$ (you can use complex exponential trick to evaluate the integral) and let $$\xi = {\sin ^{ - 1}}\left( {\frac{{{e^{ - \pi }}(1 + {e^\pi })}}{2}} \right) \in \left[ {0,\pi } \right].$$