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Solution
Your input: calculate $$$\int_{0}^{1}\left( x \cos{\left(\pi n x \right)} \right)dx$$$
First, calculate the corresponding indefinite integral: $$$\int{x \cos{\left(\pi n x \right)} d x}=\frac{\pi n x \sin{\left(\pi n x \right)} + \cos{\left(\pi n x \right)}}{\pi^{2} n^{2}}$$$ (for steps, see indefinite integral calculator)
According to the Fundamental Theorem of Calculus, $$$\int_a^b F(x) dx=f(b)-f(a)$$$, so just evaluate the integral at the endpoints, and that's the answer.
$$$\left(\frac{\pi n x \sin{\left(\pi n x \right)} + \cos{\left(\pi n x \right)}}{\pi^{2} n^{2}}\right)|_{\left(x=1\right)}=\frac{\pi n \sin{\left(\pi n \right)} + \cos{\left(\pi n \right)}}{\pi^{2} n^{2}}$$$
$$$\left(\frac{\pi n x \sin{\left(\pi n x \right)} + \cos{\left(\pi n x \right)}}{\pi^{2} n^{2}}\right)|_{\left(x=0\right)}=\frac{1}{\pi^{2} n^{2}}$$$
$$$\int_{0}^{1}\left( x \cos{\left(\pi n x \right)} \right)dx=\left(\frac{\pi n x \sin{\left(\pi n x \right)} + \cos{\left(\pi n x \right)}}{\pi^{2} n^{2}}\right)|_{\left(x=1\right)}-\left(\frac{\pi n x \sin{\left(\pi n x \right)} + \cos{\left(\pi n x \right)}}{\pi^{2} n^{2}}\right)|_{\left(x=0\right)}=\frac{\pi n \sin{\left(\pi n \right)} + \cos{\left(\pi n \right)}}{\pi^{2} n^{2}} - \frac{1}{\pi^{2} n^{2}}$$$
Answer: $$$\int_{0}^{1}\left( x \cos{\left(\pi n x \right)} \right)dx=\frac{\pi n \sin{\left(\pi n \right)} + \cos{\left(\pi n \right)}}{\pi^{2} n^{2}} - \frac{1}{\pi^{2} n^{2}}$$$