What is the integral of (e^{3x}sin(2x)) ?

The integral of (e^{3x}sin(2x)) is -(2e^{3x}cos(2x))/(13)+(3e^{3x}sin(2x))/(13)+C

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integral of (e^{3x}sin(2x))

\int (e^{3 x} sin (2 x)) d x

- \frac{2 e ^{3 x} cos ( 2 x )}{13} + \frac{3 e ^{3 x} sin ( 2 x )}{13} + C

Solution steps

\int e^{3 x} sin (2 x) d x

Apply Integration By Parts

= - \frac{1}{2} e^{3 x} cos (2 x) - \int - \frac{3}{2} e^{3 x} cos (2 x) d x

Take the constant out:

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

= - \frac{1}{2} e^{3 x} cos (2 x) - (- \frac{3}{2} \cdot \int e^{3 x} cos (2 x) d x)

Apply Integration By Parts

= - \frac{1}{2} e^{3 x} cos (2 x) - (- \frac{3}{2} (\frac{1}{2} e^{3 x} sin (2 x) - \int \frac{3}{2} e^{3 x} sin (2 x) d x))

Take the constant out:

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

= - \frac{1}{2} e^{3 x} cos (2 x) - (- \frac{3}{2} (\frac{1}{2} e^{3 x} sin (2 x) - \frac{3}{2} \cdot \int e^{3 x} sin (2 x) d x))

Therefore

\int e^{3 x} sin (2 x) d x = - \frac{1}{2} e^{3 x} cos (2 x) - (- \frac{3}{2} (\frac{1}{2} e^{3 x} sin (2 x) - \frac{3}{2} \cdot \int e^{3 x} sin (2 x) d x))

Isolate

\int e^{3 x} sin (2 x) d x

= - \frac{2 e ^{3 x} cos ( 2 x )}{13} + \frac{3 e ^{3 x} sin ( 2 x )}{13}

Add a constant to the solution

= - \frac{2 e ^{3 x} cos ( 2 x )}{13} + \frac{3 e ^{3 x} sin ( 2 x )}{13} + C

What is the integral of (e^{3x}sin(2x)) ?
The integral of (e^{3x}sin(2x)) is -(2e^{3x}cos(2x))/(13)+(3e^{3x}sin(2x))/(13)+C