What is the integral of e^{3x}sin(4x) ?

The integral of e^{3x}sin(4x) is -(4e^{3x}cos(4x))/(25)+(3e^{3x}sin(4x))/(25)+C

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integral of e^{3x}sin(4x)

\int e^{3 x} sin (4 x) d x

- \frac{4 e ^{3 x} cos ( 4 x )}{25} + \frac{3 e ^{3 x} sin ( 4 x )}{25} + C

Solution steps

\int e^{3 x} sin (4 x) d x

Apply Integration By Parts

: - \frac{1}{4} e^{3 x} cos (4 x) - \int - \frac{3}{4} e^{3 x} cos (4 x) d x

= - \frac{1}{4} e^{3 x} cos (4 x) - \int - \frac{3}{4} e^{3 x} cos (4 x) d x

Take the constant out:

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

= - \frac{1}{4} e^{3 x} cos (4 x) - (- \frac{3}{4} \cdot \int e^{3 x} cos (4 x) d x)

Apply Integration By Parts

: \frac{1}{4} e^{3 x} sin (4 x) - \int \frac{3}{4} e^{3 x} sin (4 x) d x

= - \frac{1}{4} e^{3 x} cos (4 x) - (- \frac{3}{4} (\frac{1}{4} e^{3 x} sin (4 x) - \int \frac{3}{4} e^{3 x} sin (4 x) d x))

Take the constant out:

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

= - \frac{1}{4} e^{3 x} cos (4 x) - (- \frac{3}{4} (\frac{1}{4} e^{3 x} sin (4 x) - \frac{3}{4} \cdot \int e^{3 x} sin (4 x) d x))

Therefore

\int e^{3 x} sin (4 x) d x = - \frac{1}{4} e^{3 x} cos (4 x) - (- \frac{3}{4} (\frac{1}{4} e^{3 x} sin (4 x) - \frac{3}{4} \cdot \int e^{3 x} sin (4 x) d x))

Isolate

\int e^{3 x} sin (4 x) d x

= - \frac{4 e ^{3 x} cos ( 4 x )}{25} + \frac{3 e ^{3 x} sin ( 4 x )}{25}

Add a constant to the solution

= - \frac{4 e ^{3 x} cos ( 4 x )}{25} + \frac{3 e ^{3 x} sin ( 4 x )}{25} + C

f (x) = \frac{3}{x ^{2}} - \frac{1}{x}

1 + 0.1 e^{- 0.02 t}

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = 4 x^{2}

\frac{2 s + 3}{s ^{2} + 2 s + 5}

csc^{7} (2 t^{π} + π^{3})

What is the integral of e^{3x}sin(4x) ?
The integral of e^{3x}sin(4x) is -(4e^{3x}cos(4x))/(25)+(3e^{3x}sin(4x))/(25)+C