What is the integral of sin(w)e^w ?

The integral of sin(w)e^w is -(e^wcos(w))/2+(e^wsin(w))/2+C

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integral of sin(w)e^w

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

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

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

What is the integral of sin(w)e^w ?
The integral of sin(w)e^w is -(e^wcos(w))/2+(e^wsin(w))/2+C