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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\int_{1}^{12} \frac{1}{x - 9} d x

n = 1 \sum \infty \frac{1}{n + 4}

\frac{\partial}{\partial x} (x \cdot e^{x} + ln (x) y)

\int - 7 ln (x^{2} - 1) d x

n = 2 \sum \infty \frac{1}{n ( n - 1 )}

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