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

The integral of sin(2x)e^x is -(2e^xcos(2x))/5+(e^xsin(2x))/5+C

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integral of sin(2x)e^x

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

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

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

p (x) = x^{4} - 3 x^{3} + 10 x^{2} - 6 x - 20

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

f (x) = 9 e^{x} + x, at x = 0

\int - e^{- 2 t} d t

\frac{\partial}{\partial x} (\frac{1}{2 π t} e^{- \frac{( x - y ) ^{2}}{2 t}})

What is the integral of sin(2x)e^x ?
The integral of sin(2x)e^x is -(2e^xcos(2x))/5+(e^xsin(2x))/5+C