What is the integral of e^xsin(npix) ?

The integral of e^xsin(npix) is -(pine^xcos(pinx))/(pi^2n^2+1)+(e^xsin(pinx))/(pi^2n^2+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^xsin(npix)

\int e^{x} sin (nπ x) d x

- \frac{πn e ^{x} cos ( πn x )}{π ^{2} n ^{2} + 1} + \frac{e ^{x} sin ( πn x )}{π ^{2} n ^{2} + 1} + C

Solution steps

\int e^{x} sin (nπ x) d x

Apply Integration By Parts

= - e^{x} \frac{1}{πn} cos (πn x) - \int - e^{x} \frac{1}{πn} cos (πn x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - e^{x} \frac{1}{πn} cos (πn x) - (- \frac{1}{πn} \cdot \int e^{x} cos (πn x) d x)

Apply Integration By Parts

= - e^{x} \frac{1}{πn} cos (πn x) - (- \frac{1}{πn} (e^{x} \frac{1}{πn} sin (πn x) - \int e^{x} \frac{1}{πn} sin (πn x) d x))

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - e^{x} \frac{1}{πn} cos (πn x) - (- \frac{1}{πn} (e^{x} \frac{1}{πn} sin (πn x) - \frac{1}{πn} \cdot \int e^{x} sin (πn x) d x))

Therefore

\int e^{x} sin (πn x) d x = - e^{x} \frac{1}{πn} cos (πn x) - (- \frac{1}{πn} (e^{x} \frac{1}{πn} sin (πn x) - \frac{1}{πn} \cdot \int e^{x} sin (πn x) d x))

Isolate

\int e^{x} sin (πn x) d x

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

Add a constant to the solution

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

What is the integral of e^xsin(npix) ?
The integral of e^xsin(npix) is -(pine^xcos(pinx))/(pi^2n^2+1)+(e^xsin(pinx))/(pi^2n^2+1)+C