Does the integral $\int_0^{\pi} \frac{dx}{\sin(2x)+\cos(3x)}$ exist? This link
http://www.wolframalpha.com/input/?i=integral%28x%3D0%2Cpi%2C1%2F%28sin2x%2Bcos3x%29%29
shows the visual representation of the integral
$$\int_0^{\pi} \frac{dx}{\sin(2x)+\cos(3x)}$$
Looking at the picture, I tend to believe that the integral does not exist.
The integrand contains three poles, but from this I cannot conclude that the
integral does not exist, or can I ?


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*How can I check, if the integral exists or not ?


Note at $x = {\pi \over 2}$, the denominator $\sin 2x + \cos 3x$ is zero. If $f(x) = \sin 2x + \cos 3x$, then $f'(x) = 2 \cos 2x - 3 \sin 3x$, so that $f'({\pi \over 2}) = 1$. Hence the function $f(x)$ behaves as ${1 \over x - {\pi \over 2}}$ near $x = {\pi \over 2}$; in particular, since $|{1 \over x - {\pi \over 2}}|$ has infinite integral in any interval centered at ${\pi \over 2}$, so does $|{1 \over \sin 2x + \cos 3x}|$. Thus in the usual sense your integral does not exist.
There are more general notions of integral where you remove intervals of the form $({\pi \over 2} - \epsilon, {\pi \over 2} + \epsilon)$ (and analogous things for the other poles), and then take limits as $\epsilon \rightarrow 0$. If you do this, sometimes you can get a finite number as your limit for integrals such as yours.
