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) \cdot 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

\int_{0}^{\frac{1}{5}} 2 arcsin (5 y) d y

f (x) = sin (x) csc (x)

(sec (x) + tan (x)) (sec (x) - tan (x))

\int \frac{1}{64 e ^{- 3 x} + e ^{3 x}} d x

n = 0 \sum \infty sin (nπ)

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