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

The integral of e^xsin(x/2) is -2/5 e^xcos(x/2)+4/5 e^xsin(x/2)+C

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integral of e^xsin(x/2)

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

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

Solution steps

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

Apply u-substitution

: \int e^{2 u} sin (u) \cdot 2 d u

= \int e^{2 u} sin (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int e^{2 u} sin (u) d u

Apply Integration By Parts

: - e^{2 u} cos (u) - \int - 2 e^{2 u} cos (u) d u

= 2 (- e^{2 u} cos (u) - \int - 2 e^{2 u} cos (u) d u)

Take the constant out:

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

= 2 (- e^{2 u} cos (u) - (- 2 \cdot \int e^{2 u} cos (u) d u))

Apply Integration By Parts

: e^{2 u} sin (u) - \int e^{2 u} \cdot 2 sin (u) d u

= 2 (- e^{2 u} cos (u) - (- 2 (e^{2 u} sin (u) - \int e^{2 u} \cdot 2 sin (u) d u)))

Therefore

\int e^{2 u} \cdot 2 sin (u) d u = 2 (- e^{2 u} cos (u) - (- 2 (e^{2 u} sin (u) - \int e^{2 u} \cdot 2 sin (u) d u)))

Isolate

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

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} ((x^{2} + 1) x)

\frac{d}{d x} ((a x^{2} + b x + c) \cdot e^{- \frac{x}{2}})

4 v^{4} - (12 v^{3} + 9 v)

\frac{d y}{d x} = \frac{( x - y )}{x + y}

\frac{d}{d x} (4 + \frac{4}{x})

What is the integral of e^xsin(x/2) ?
The integral of e^xsin(x/2) is -2/5 e^xcos(x/2)+4/5 e^xsin(x/2)+C