What is the integral of e^zsin(z) ?

The integral of e^zsin(z) is -(e^zcos(z))/2+(e^zsin(z))/2+C

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integral of e^zsin(z)

\int e^{z} sin (z) d z

- \frac{e ^{z} cos ( z )}{2} + \frac{e ^{z} sin ( z )}{2} + C

Solution steps

\int e^{z} sin (z) d z

Apply Integration By Parts

: - e^{z} cos (z) - \int - e^{z} cos (z) d z

= - e^{z} cos (z) - \int - e^{z} cos (z) d z

Take the constant out:

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

= - e^{z} cos (z) - (- \int e^{z} cos (z) d z)

Apply Integration By Parts

: e^{z} sin (z) - \int e^{z} sin (z) d z

= - e^{z} cos (z) - (- (e^{z} sin (z) - \int e^{z} sin (z) d z))

Therefore

\int e^{z} sin (z) d z = - e^{z} cos (z) - (- (e^{z} sin (z) - \int e^{z} sin (z) d z))

Isolate

\int e^{z} sin (z) d z

= - \frac{e ^{z} cos ( z )}{2} + \frac{e ^{z} sin ( z )}{2}

Add a constant to the solution

= - \frac{e ^{z} cos ( z )}{2} + \frac{e ^{z} sin ( z )}{2} + C

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What is the integral of e^zsin(z) ?
The integral of e^zsin(z) is -(e^zcos(z))/2+(e^zsin(z))/2+C