91Ó°ÊÓ

If f(x) is an even function, that means \(f(-x)=f(x)\). We need to determine the properties of \(\int_{-\pi}^{\pi}f(x)cos(nx)dx\) and \(\int_{-\pi}^{\pi}f(x)sin(nx)dx\). Now, let's examine two cases: 1) For \(\int_{-\pi}^{\pi}f(x)cos(nx)dx\): We can use the substitution \(u=-x\). Therefore, \(du=-dx\), and we get: \[\int_{\pi}^{-\pi}f(-u)cos(-nu)(-du) = \int_{-\pi}^{\pi}f(u)cos(nu)du\] Since the limits of integration are the same, we can denote the equation as: \[\int_{-\pi}^{\pi}f(x)cos(nx)dx = \int_{-\pi}^{\pi}f(-x)cos(nx)dx = \int_{-\pi}^{\pi}f(x)cos(nx)dx\] Thus, for even functions, the integral does not change when integrated with cosine functions. 2) For \(\int_{-\pi}^{\pi}f(x)sin(nx)dx\): Again, we use the substitution \(u=-x\), so \[\int_{\pi}^{-\pi}f(-u)sin(-nu)(-du) = \int_{-\pi}^{\pi}f(u)sin(nu)du\] We have \(\int_{-\pi}^{\pi}f(x)sin(nx)dx + \int_{-\pi}^{\pi}f(x)sin(nx)dx = 0\), which means \[\int_{-\pi}^{\pi}f(x)sin(nx)dx = 0\] As a result, for an even function \(f(x)\), \(\int_{-\pi}^{\pi}f(x)cos(nx)dx\) remains the same, and \(\int_{-\pi}^{\pi}f(x)sin(nx)dx = 0\).

If f(x) is an odd function, that means \(f(-x)=-f(x)\). Now we'll analyze the properties of \(\int_{-\pi}^{\pi}f(x)cos(nx)dx\) and \(\int_{-\pi}^{\pi}f(x)sin(nx)dx\). 1) For \(\int_{-\pi}^{\pi}f(x)cos(nx)dx\): Using the same substitution, \(u=-x\), we get: \[\int_{-\pi}^{\pi}f(-x)cos(nx)dx = -\int_{-\pi}^{\pi}f(x)cos(nx)dx\] Thus, \(\int_{-\pi}^{\pi}f(x)cos(nx)dx + \int_{-\pi}^{\pi}f(x)cos(nx)dx = 0\), and we obtain: \[\int_{-\pi}^{\pi}f(x)cos(nx)dx = 0\] 2) For \(\int_{-\pi}^{\pi}f(x)sin(nx)dx\): With the same substitution, we have: \[\int_{-\pi}^{\pi}f(-x)sin(nx)dx = \int_{-\pi}^{\pi}(-f(x))sin(nx)dx = -\int_{-\pi}^{\pi}f(x)sin(nx)dx\] This implies that, for odd functions, \(\int_{-\pi}^{\pi}f(x)sin(nx)dx\) remains the same. In conclusion, for an odd function \(f(x)\), \(\int_{-\pi}^{\pi}f(x)cos(nx)dx = 0\), while \(\int_{-\pi}^{\pi}f(x)sin(nx)dx\) remains the same.