What is the integral of e^{4x}sin(5x) ?

The integral of e^{4x}sin(5x) is -(5e^{4x}cos(5x))/(41)+(4e^{4x}sin(5x))/(41)+C

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integral of e^{4x}sin(5x)

\int e^{4 x} sin (5 x) d x

- \frac{5 e ^{4 x} cos ( 5 x )}{41} + \frac{4 e ^{4 x} sin ( 5 x )}{41} + C

Solution steps

\int e^{4 x} sin (5 x) d x

Apply Integration By Parts

= - \frac{1}{5} e^{4 x} cos (5 x) - \int - \frac{4}{5} e^{4 x} cos (5 x) d x

Take the constant out:

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

= - \frac{1}{5} e^{4 x} cos (5 x) - (- \frac{4}{5} \cdot \int e^{4 x} cos (5 x) d x)

Apply Integration By Parts

= - \frac{1}{5} e^{4 x} cos (5 x) - (- \frac{4}{5} (\frac{1}{5} e^{4 x} sin (5 x) - \int \frac{4}{5} e^{4 x} sin (5 x) d x))

Take the constant out:

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

= - \frac{1}{5} e^{4 x} cos (5 x) - (- \frac{4}{5} (\frac{1}{5} e^{4 x} sin (5 x) - \frac{4}{5} \cdot \int e^{4 x} sin (5 x) d x))

Therefore

\int e^{4 x} sin (5 x) d x = - \frac{1}{5} e^{4 x} cos (5 x) - (- \frac{4}{5} (\frac{1}{5} e^{4 x} sin (5 x) - \frac{4}{5} \cdot \int e^{4 x} sin (5 x) d x))

Isolate

\int e^{4 x} sin (5 x) d x

= - \frac{5 e ^{4 x} cos ( 5 x )}{41} + \frac{4 e ^{4 x} sin ( 5 x )}{41}

Add a constant to the solution

= - \frac{5 e ^{4 x} cos ( 5 x )}{41} + \frac{4 e ^{4 x} sin ( 5 x )}{41} + C

x \to 0 lim (\frac{4}{x} - \frac{4}{x ^{2} + x})

\frac{d y}{d x} = \frac{( 2 x y )}{x ^{2} - y ^{2}}

\frac{d}{d a} (ab c d + 7 ab - 2 b c + 5 c d - 5)

\frac{d y}{d x} + 1 = 2 y, y (1) = 1

\int \frac{2 u - 1}{4 u ^{2} - 1} d u

What is the integral of e^{4x}sin(5x) ?
The integral of e^{4x}sin(5x) is -(5e^{4x}cos(5x))/(41)+(4e^{4x}sin(5x))/(41)+C