What is the integral of e^{(3-z)}sin(2+z) ?

The integral of e^{(3-z)}sin(2+z) is -(e^{3-z}cos(2+z))/2-(e^{3-z}sin(2+z))/2+C

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integral of e^{(3-z)}sin(2+z)

\int e^{(3 - z)} sin (2 + z) d z

- \frac{e ^{3 - z} cos ( 2 + z )}{2} - \frac{e ^{3 - z} sin ( 2 + z )}{2} + C

Solution steps

\int e^{3 - z} sin (2 + z) d z

Apply Integration By Parts

: - e^{3 - z} cos (2 + z) - \int e^{3 - z} cos (2 + z) d z

= - e^{3 - z} cos (2 + z) - \int e^{3 - z} cos (2 + z) d z

Apply Integration By Parts

: e^{3 - z} sin (2 + z) - \int - e^{3 - z} sin (2 + z) d z

= - e^{3 - z} cos (2 + z) - (e^{3 - z} sin (2 + z) - \int - e^{3 - z} sin (2 + z) d z)

Take the constant out:

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

= - e^{3 - z} cos (2 + z) - (e^{3 - z} sin (2 + z) - (- \int e^{3 - z} sin (2 + z) d z))

Therefore

\int e^{3 - z} sin (2 + z) d z = - e^{3 - z} cos (2 + z) - (e^{3 - z} sin (2 + z) - (- \int e^{3 - z} sin (2 + z) d z))

Isolate

\int e^{3 - z} sin (2 + z) d z

= - \frac{e ^{3 - z} cos ( 2 + z )}{2} - \frac{e ^{3 - z} sin ( 2 + z )}{2}

Add a constant to the solution

= - \frac{e ^{3 - z} cos ( 2 + z )}{2} - \frac{e ^{3 - z} sin ( 2 + z )}{2} + C

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What is the integral of e^{(3-z)}sin(2+z) ?
The integral of e^{(3-z)}sin(2+z) is -(e^{3-z}cos(2+z))/2-(e^{3-z}sin(2+z))/2+C