What is the integral of sin(2x)e^{5x} ?

The integral of sin(2x)e^{5x} is -(2e^{5x}cos(2x))/(29)+(5e^{5x}sin(2x))/(29)+C

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integral of sin(2x)e^{5x}

\int sin (2 x) e^{5 x} d x

- \frac{2 e ^{5 x} cos ( 2 x )}{29} + \frac{5 e ^{5 x} sin ( 2 x )}{29} + C

Solution steps

\int sin (2 x) e^{5 x} d x

Apply Integration By Parts

= - \frac{1}{2} e^{5 x} cos (2 x) - \int - \frac{5}{2} e^{5 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^{5 x} cos (2 x) - (- \frac{5}{2} \cdot \int e^{5 x} cos (2 x) d x)

Apply Integration By Parts

= - \frac{1}{2} e^{5 x} cos (2 x) - (- \frac{5}{2} (\frac{1}{2} e^{5 x} sin (2 x) - \int \frac{5}{2} e^{5 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^{5 x} cos (2 x) - (- \frac{5}{2} (\frac{1}{2} e^{5 x} sin (2 x) - \frac{5}{2} \cdot \int e^{5 x} sin (2 x) d x))

Therefore

\int e^{5 x} sin (2 x) d x = - \frac{1}{2} e^{5 x} cos (2 x) - (- \frac{5}{2} (\frac{1}{2} e^{5 x} sin (2 x) - \frac{5}{2} \cdot \int e^{5 x} sin (2 x) d x))

Isolate

\int e^{5 x} sin (2 x) d x

= - \frac{2 e ^{5 x} cos ( 2 x )}{29} + \frac{5 e ^{5 x} sin ( 2 x )}{29}

Add a constant to the solution

= - \frac{2 e ^{5 x} cos ( 2 x )}{29} + \frac{5 e ^{5 x} sin ( 2 x )}{29} + C

What is the integral of sin(2x)e^{5x} ?
The integral of sin(2x)e^{5x} is -(2e^{5x}cos(2x))/(29)+(5e^{5x}sin(2x))/(29)+C