为什么不整齐
$$\int_{0}^{2\pi}\sin xdx=[-\cos x]_{0}^{2\pi}\\ = \cos2\pi-\cos0\\ = 1-1\\ = 0$$
∫02πsinxdx=[−cosx]02π=cos2π−cos0=1−1=0\int_{0}^{2\pi}\sin xdx=[-\cos x]_{0}^{2\pi}\\ = \cos2\pi-\cos0\\ = 1-1\\ = 0∫02πsinxdx=[−cosx]02π=cos2π−cos0=1−1=0
