integral from 0 to 2pi of e^{i(θ-x)}

解答

\int_{0}^{2 π} e^{i (θ - x)} d x

- sin (θ - 2 π) + sin (θ) + i (- cos (θ) + cos (θ - 2 π))

求解步骤

\int_{0}^{2 π} e^{i (θ - x)} d x

使用虚数运算法则:

e^{ia} = cos (a) + i sin (a)

= \int_{0}^{2 π} cos (θ - x) + i sin (θ - x) d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2 π} cos (θ - x) d x + \int_{0}^{2 π} i sin (θ - x) d x

\int_{0}^{2 π} cos (θ - x) d x = - sin (θ - 2 π) + sin (θ)

\int_{0}^{2 π} i sin (θ - x) d x = i (- cos (θ) + cos (θ - 2 π))

= - sin (θ - 2 π) + sin (θ) + i (- cos (θ) + cos (θ - 2 π))

\int \frac{4 x ^{2}}{4 x - x ^{2}} d x

\frac{d}{d x} (8 x e^{2 x})

\int e^{x} tan^{4} (e^{x}) d x

n = 1 \sum \infty \frac{n}{n + 1}

\int - t^{3} d t