What is the integral of e^{-x}sin(2x+1) ?

The integral of e^{-x}sin(2x+1) is -(2e^{-x}cos(2x+1))/5-(e^{-x}sin(2x+1))/5+C

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integral of e^{-x}sin(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

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

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

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

\frac{1}{x ^{2} - 1}

\int - cos (x) 2 d x

\int (x - 5) 2 x d x

\int \frac{s}{s + 1} d s

What is the integral of e^{-x}sin(2x+1) ?
The integral of e^{-x}sin(2x+1) is -(2e^{-x}cos(2x+1))/5-(e^{-x}sin(2x+1))/5+C