integral from 0 to t of e^xsin(pix)

解答

\int_{0}^{t} e^{x} \cdot sin (π \cdot x) d x

\frac{- π e ^{t} cos ( π t ) + e ^{t} sin ( π t ) + π}{π ^{2} + 1}

求解步骤

\int_{0}^{t} e^{x} sin (π x) d x

计算不定积分:

\int e^{x} sin (π x) d x = - \frac{π e ^{x} cos ( π x )}{π ^{2} + 1} + \frac{e ^{x} sin ( π x )}{π ^{2} + 1} + C

带入上下限计算后得:

\int_{0}^{t} e^{x} sin (π x) d x = - \frac{π e ^{t} cos ( π t )}{π ^{2} + 1} + \frac{e ^{t} sin ( π t )}{π ^{2} + 1} - (- \frac{π}{π ^{2} + 1})

= - \frac{π e ^{t} cos ( π t )}{π ^{2} + 1} + \frac{e ^{t} sin ( π t )}{π ^{2} + 1} - (- \frac{π}{π ^{2} + 1})

化简

= \frac{- π e ^{t} cos ( π t ) + e ^{t} sin ( π t ) + π}{π ^{2} + 1}

t an g e n t 6 e^{x} cos (x)

x \to 6 lim (x^{2} - 36)

\frac{\partial}{\partial x} (5 x)

x \to 0 lim (\frac{( tan ( 5 x ) )}{x})

\int 12 x^{2} ln (x) d x