What is the integral of sin(x)e^{x+1} ?

The integral of sin(x)e^{x+1} is -(e^{x+1}cos(x))/2+(e^{x+1}sin(x))/2+C

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integral of sin(x)e^{x+1}

\int sin (x) e^{x + 1} d x

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

Solution steps

\int sin (x) e^{x + 1} d x

Apply Integration By Parts

: - e^{x + 1} cos (x) - \int - e^{x + 1} cos (x) d x

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

Take the constant out:

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

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

Apply Integration By Parts

: e^{x + 1} sin (x) - \int e^{x + 1} sin (x) d x

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

Therefore

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

Isolate

\int e^{x + 1} sin (x) d x

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

Add a constant to the solution

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

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What is the integral of sin(x)e^{x+1} ?
The integral of sin(x)e^{x+1} is -(e^{x+1}cos(x))/2+(e^{x+1}sin(x))/2+C