What is the integral of sin(3x)e^{3x+1} ?

The integral of sin(3x)e^{3x+1} is -(e^{3x+1}cos(3x))/6+(e^{3x+1}sin(3x))/6+C

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integral of sin(3x)e^{3x+1}

\int sin (3 x) e^{3 x + 1} d x

- \frac{e ^{3 x + 1} cos ( 3 x )}{6} + \frac{e ^{3 x + 1} sin ( 3 x )}{6} + C

Solution steps

\int sin (3 x) e^{3 x + 1} d x

Apply Integration By Parts

: - \frac{1}{3} e^{3 x + 1} cos (3 x) - \int - e^{3 x + 1} cos (3 x) d x

= - \frac{1}{3} e^{3 x + 1} cos (3 x) - \int - e^{3 x + 1} cos (3 x) d x

Take the constant out:

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

= - \frac{1}{3} e^{3 x + 1} cos (3 x) - (- \int e^{3 x + 1} cos (3 x) d x)

Apply Integration By Parts

: \frac{1}{3} e^{3 x + 1} sin (3 x) - \int e^{3 x + 1} sin (3 x) d x

= - \frac{1}{3} e^{3 x + 1} cos (3 x) - (- (\frac{1}{3} e^{3 x + 1} sin (3 x) - \int e^{3 x + 1} sin (3 x) d x))

Therefore

\int e^{3 x + 1} sin (3 x) d x = - \frac{1}{3} e^{3 x + 1} cos (3 x) - (- (\frac{1}{3} e^{3 x + 1} sin (3 x) - \int e^{3 x + 1} sin (3 x) d x))

Isolate

\int e^{3 x + 1} sin (3 x) d x

= - \frac{e ^{3 x + 1} cos ( 3 x )}{6} + \frac{e ^{3 x + 1} sin ( 3 x )}{6}

Add a constant to the solution

= - \frac{e ^{3 x + 1} cos ( 3 x )}{6} + \frac{e ^{3 x + 1} sin ( 3 x )}{6} + C

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What is the integral of sin(3x)e^{3x+1} ?
The integral of sin(3x)e^{3x+1} is -(e^{3x+1}cos(3x))/6+(e^{3x+1}sin(3x))/6+C