$\int_{c}^{c+2L} cos(\frac{2\pi nx}{L}) = \frac{L sin(2\pi n) cos(\frac{2 \pi n(c+L)}{L})}{\pi n}$ dx with $n = 0,1,2,...$
I put this integral in Wolfram Alpha and gives me this result, but when I try to do it at hand I arrive to other result. Is Wolfram Alpha wrong?
Upon integrating, we have for $n\ne 0$, the term
$$\begin{align} \sin \left(\frac{2\pi n(c+2L)}{L}\right)-\sin \left(\frac{2\pi nc}{L}\right)&=\sin \left(\frac{2\pi nc}{L}\right)\cos(4\pi n)\\\\ &+\cos \left(\frac{2\pi nc}{L}\right)\sin(4\pi n)\\\\ &-\sin \left(\frac{2\pi nc}{L}\right)\\\\ &=\sin \left(\frac{2\pi nc}{L}\right)\left(cos(4\pi n)-1\right)\\\\ &+\cos \left(\frac{2\pi nc}{L}\right)\sin(4\pi n)\\\\ &=\sin \left(\frac{2\pi nc}{L}\right)\left(-2\sin^2(2\pi n)\right)\\\\ &+\cos \left(\frac{2\pi nc}{L}\right)\left(2\sin(2\pi n)\cos(2\pi n)\right)\\\\ &=2\sin(2\pi n)\cos \left(\frac{2\pi nc}{L}+2\pi n\right)\\\\ &=2\sin(2\pi n)\cos \left(\frac{2\pi n(C+L)}{L}\right)\\\\ \end{align}$$
Therefore, the integral is for $n\ne 0$
$$\int_c^{c+2L}\cos (2\pi nx/L)\,dx= \begin{cases} \frac{L\sin(2\pi n)\cos \left(\frac{2\pi n(C+L)}{L}\right)}{\pi n}&,n\ne 0\\\\ 2L&,n=0 \end{cases}$$
If $n$ is an integer, then we have
$$\int_c^{c+2L}\cos (2\pi nx/L)\,dx= \begin{cases} 0&,n\ne 0\\\\ 2L&,n=0 \end{cases}$$
No, Wolfram Alpha is correct. But since your $n$ is an integer (which of course Wolfram Alpha does not assume if you don't say so), $\sin(2\pi n)=0$. Hence, the result is zero.
Edit
I suggest that you take care of the integral for $n=0$ by hand. Wolfram Alpha gives a generic result...
