integral of e^{-iax}

解答

\int e^{- ia x} d x

\frac{1}{a} sin (a x) + i \frac{1}{a} cos (a x) + C

求解步骤

\int e^{- ia x} d x

化简

e^{- ia x} : cos (a x) - i sin (a x)

= \int cos (a x) - i sin (a 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 cos (a x) d x - \int i sin (a x) d x

\int cos (a x) d x = \frac{1}{a} sin (a x)

\int i sin (a x) d x = - i \frac{1}{a} cos (a x)

= \frac{1}{a} sin (a x) - (- i \frac{1}{a} cos (a x))

化简

\frac{1}{a} sin (a x) - (- i \frac{1}{a} cos (a x)) : \frac{1}{a} sin (a x) + i \frac{1}{a} cos (a x)

= \frac{1}{a} sin (a x) + i \frac{1}{a} cos (a x)

解答补常数

= \frac{1}{a} sin (a x) + i \frac{1}{a} cos (a x) + C

\int 17 tan^{3} (θ) d θ

\int ln (e^{- x^{2}}) d x

a re a 5 x^{4} - 5 x^{2}, 9 x^{2}

\int \frac{x ^{2}}{9 e ^{3 x}} d x

d er i v a t i v e y = \frac{8 x}{5 - tan ( x )}