What is the integral of e^xsin(2x+1) ?

The integral of e^xsin(2x+1) is -(2e^xcos(2x+1))/5+(e^xsin(2x+1))/5+C

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integral of e^xsin(2x+1)

\int e^{x} sin (2 x + 1) d x

- \frac{2 e ^{x} cos ( 2 x + 1 )}{5} + \frac{e ^{x} sin ( 2 x + 1 )}{5} + C

Solution steps

\int e^{x} sin (2 x + 1) d x

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

\int e^{x} sin (2 x + 1) d x

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

Add a constant to the solution

= - \frac{2 e ^{x} cos ( 2 x + 1 )}{5} + \frac{e ^{x} sin ( 2 x + 1 )}{5} + C

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What is the integral of e^xsin(2x+1) ?
The integral of e^xsin(2x+1) is -(2e^xcos(2x+1))/5+(e^xsin(2x+1))/5+C