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

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

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integral of e^{x/2}cos(x)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Substitute back

u = \frac{x}{2}

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

Simplify

2 (\frac{2 e ^{\frac{x}{2}} sin ( 2 \cdot \frac{x}{2} )}{5} + \frac{e ^{\frac{x}{2}} cos ( 2 \cdot \frac{x}{2} )}{5}) : 2 (\frac{2}{5} e^{\frac{x}{2}} sin (x) + \frac{1}{5} e^{\frac{x}{2}} cos (x))

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

Add a constant to the solution

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

\frac{d}{d x} (8 e^{x} + x e^{x})

9 (t + 1) y^{^{'}} - 5 y = 20 t

\frac{\partial}{\partial s} (ln (r^{2} + s^{2} + 2))

\int e^{x} 25 - e^{2 x} d x

f (x) = x \cdot e^{x}

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