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

The integral of e^{-x}sin(x/2) is -(2e^{-x}cos(1/2 x))/5-(4e^{-x}sin(1/2 x))/5+C

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integral of e^{-x}sin(x/2)

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

- \frac{2 e ^{- x} cos ( \frac{1}{2} x )}{5} - \frac{4 e ^{- x} sin ( \frac{1}{2} x )}{5} + C

Solution steps

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

\frac{x}{2} = x \frac{1}{2}

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

Apply Integration By Parts

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Therefore

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

Isolate

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

= - \frac{2 e ^{- x} cos ( \frac{1}{2} x )}{5} - \frac{4 e ^{- x} sin ( \frac{1}{2} x )}{5}

Add a constant to the solution

= - \frac{2 e ^{- x} cos ( \frac{1}{2} x )}{5} - \frac{4 e ^{- x} sin ( \frac{1}{2} x )}{5} + C

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What is the integral of e^{-x}sin(x/2) ?
The integral of e^{-x}sin(x/2) is -(2e^{-x}cos(1/2 x))/5-(4e^{-x}sin(1/2 x))/5+C