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

= - \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} sin (2 x) - \int \frac{3}{2} 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) - \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

n = 3 \sum \infty e^{3 - 6 n}

\frac{\partial}{\partial x} (e^{x + y} - 1)

\int \frac{e ^{- 2 x}}{e ^{x}} d x

\frac{d}{d x} (4 x^{\frac{5}{2}})

x \to - 2 lim (\frac{2 + ∣ x ∣}{2 + x})

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