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

The integral of sin(u)e^u is -(e^ucos(u))/2+(e^usin(u))/2+C

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integral of sin(u)e^u

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

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

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

What is the integral of sin(u)e^u ?
The integral of sin(u)e^u is -(e^ucos(u))/2+(e^usin(u))/2+C