What is the integral of e^{4x}cos(x/2) ?

The integral of e^{4x}cos(x/2) is 2(1/65 e^{4x}sin(x/2)+8/65 e^{4x}cos(x/2))+C

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integral of e^{4x}cos(x/2)

\int e^{4 x} cos (\frac{x}{2}) d x

2 (\frac{1}{65} e^{4 x} sin (\frac{x}{2}) + \frac{8}{65} e^{4 x} cos (\frac{x}{2})) + C

Solution steps

\int e^{4 x} cos (\frac{x}{2}) d x

Apply u-substitution

: \int 2 e^{8 u} cos (u) d u

= \int 2 e^{8 u} cos (u) d u

Take the constant out:

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

= 2 \cdot \int e^{8 u} cos (u) d u

Apply Integration By Parts

: e^{8 u} sin (u) - \int e^{8 u} \cdot 8 sin (u) d u

= 2 (e^{8 u} sin (u) - \int e^{8 u} \cdot 8 sin (u) d u)

Take the constant out:

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

= 2 (e^{8 u} sin (u) - 8 \cdot \int e^{8 u} sin (u) d u)

Apply Integration By Parts

: - e^{8 u} cos (u) - \int - 8 e^{8 u} cos (u) d u

= 2 (e^{8 u} sin (u) - 8 (- e^{8 u} cos (u) - \int - 8 e^{8 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^{8 u} sin (u) - 8 (- e^{8 u} cos (u) - (- 8 \cdot \int e^{8 u} cos (u) d u)))

Therefore

2 \cdot \int e^{8 u} cos (u) d u = 2 (e^{8 u} sin (u) - 8 (- e^{8 u} cos (u) - (- 8 \cdot \int e^{8 u} cos (u) d u)))

Isolate

\int e^{8 u} cos (u) d u

= 2 (\frac{e ^{8 u} sin ( u )}{65} + \frac{8 e ^{8 u} cos ( u )}{65})

Substitute back

u = \frac{x}{2}

= 2 (\frac{e ^{8 \cdot \frac{x}{2}} sin ( \frac{x}{2} )}{65} + \frac{8 e ^{8 \cdot \frac{x}{2}} cos ( \frac{x}{2} )}{65})

Simplify

2 (\frac{e ^{8 \cdot \frac{x}{2}} sin ( \frac{x}{2} )}{65} + \frac{8 e ^{8 \cdot \frac{x}{2}} cos ( \frac{x}{2} )}{65}) : 2 (\frac{1}{65} e^{4 x} sin (\frac{x}{2}) + \frac{8}{65} e^{4 x} cos (\frac{x}{2}))

= 2 (\frac{1}{65} e^{4 x} sin (\frac{x}{2}) + \frac{8}{65} e^{4 x} cos (\frac{x}{2}))

Add a constant to the solution

= 2 (\frac{1}{65} e^{4 x} sin (\frac{x}{2}) + \frac{8}{65} e^{4 x} cos (\frac{x}{2})) + C

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What is the integral of e^{4x}cos(x/2) ?
The integral of e^{4x}cos(x/2) is 2(1/65 e^{4x}sin(x/2)+8/65 e^{4x}cos(x/2))+C