Have been asked the question find the fourier series $f(x) = x^4, x \in (−4,4)$.
not sure if its correct but u have
$$A_n = \frac{1}{4} \int_{-4}^4 x^4 \cos \bigg(\frac{\pi n}{4}x\bigg) dx$$
im stuck on trying to integrate this and plug back into formula
Try to integrate by parts: $$\left|\begin{array}{c|c|c} &D & I \\ \hline + & x^4 & \cos(ax)\\ - & 4x^3 & \frac{1}{a} \sin(ax)\\ + & 12x^2 & -\frac{1}{a^2} \cos(ax)\\ - & 24x & -\frac{1}{a^3} \sin(ax)\\ + & 24 & \frac{1}{a^4} \cos(ax)\\ - & 0 & \frac{1}{a^5} \sin(ax)\\ \end{array}\right|$$ So : $$\int x^4 \cos(ax) \mathrm{d}x=x^4 \frac{1}{a} \sin(ax) - 4x^3 \left(-\frac{1}{a^2} \cos(ax)\right)+12x^2 \left(-\frac{1}{a^3} \sin(ax)\right)-24x \frac{1}{a^4} \cos(ax)+24\frac{1}{a^5} \sin(ax)$$ Substituting back the $a=\frac{n \pi}{4}$: $$x^4 \frac{4}{n \pi}\sin\left(\frac{n \pi}{4}x\right)+4x^3\frac{4^2}{n^2 \pi^2}\cos\left(\frac{n \pi}{4}x\right)-12x^2\frac{4^3}{n^3 \pi^3}\sin\left(\frac{n \pi}{4}x\right)-24x \frac{4^4}{n^4 \pi^4} \cos\left(\frac{n \pi}{4}x\right)+24 \frac{4^5}{n^5 \pi^5}\sin\left(\frac{n \pi}{4}x\right)$$ Now we can group some term: $$\left(x^4 \frac{4}{n \pi}-12x^2\frac{4^3}{n^3 \pi^3}+24 \frac{4^5}{n^5 \pi^5}\right)\sin\left(\frac{n \pi}{4}x\right) + \left(4x^3\frac{4^2}{n^2 \pi^2} -24x \frac{4^4}{n^4 \pi^4}\right)\cos\left(\frac{n \pi}{4}x\right)$$ Note: I did not write down the $+$ constant, because it will be used to calculate a definite integral.