What is the integral of e^{3x}cos(x/2) ?

The integral of e^{3x}cos(x/2) is 2(1/37 e^{3x}sin(x/2)+6/37 e^{3x}cos(x/2))+C

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integral of e^{3x}cos(x/2)

\int e^{3 x} cos (\frac{x}{2}) d x

2 (\frac{1}{37} e^{3 x} sin (\frac{x}{2}) + \frac{6}{37} e^{3 x} cos (\frac{x}{2})) + C

Solution steps

\int e^{3 x} cos (\frac{x}{2}) d x

Apply u-substitution

: \int 2 e^{6 u} cos (u) d u

= \int 2 e^{6 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^{6 u} cos (u) d u

Apply Integration By Parts

: e^{6 u} sin (u) - \int e^{6 u} \cdot 6 sin (u) d u

= 2 (e^{6 u} sin (u) - \int e^{6 u} \cdot 6 sin (u) d u)

Take the constant out:

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

= 2 (e^{6 u} sin (u) - 6 \cdot \int e^{6 u} sin (u) d u)

Apply Integration By Parts

: - e^{6 u} cos (u) - \int - 6 e^{6 u} cos (u) d u

= 2 (e^{6 u} sin (u) - 6 (- e^{6 u} cos (u) - \int - 6 e^{6 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^{6 u} sin (u) - 6 (- e^{6 u} cos (u) - (- 6 \cdot \int e^{6 u} cos (u) d u)))

Therefore

2 \cdot \int e^{6 u} cos (u) d u = 2 (e^{6 u} sin (u) - 6 (- e^{6 u} cos (u) - (- 6 \cdot \int e^{6 u} cos (u) d u)))

Isolate

\int e^{6 u} cos (u) d u

= 2 (\frac{e ^{6 u} sin ( u )}{37} + \frac{6 e ^{6 u} cos ( u )}{37})

Substitute back

u = \frac{x}{2}

= 2 (\frac{e ^{6 \cdot \frac{x}{2}} sin ( \frac{x}{2} )}{37} + \frac{6 e ^{6 \cdot \frac{x}{2}} cos ( \frac{x}{2} )}{37})

Simplify

2 (\frac{e ^{6 \cdot \frac{x}{2}} sin ( \frac{x}{2} )}{37} + \frac{6 e ^{6 \cdot \frac{x}{2}} cos ( \frac{x}{2} )}{37}) : 2 (\frac{1}{37} e^{3 x} sin (\frac{x}{2}) + \frac{6}{37} e^{3 x} cos (\frac{x}{2}))

= 2 (\frac{1}{37} e^{3 x} sin (\frac{x}{2}) + \frac{6}{37} e^{3 x} cos (\frac{x}{2}))

Add a constant to the solution

= 2 (\frac{1}{37} e^{3 x} sin (\frac{x}{2}) + \frac{6}{37} e^{3 x} cos (\frac{x}{2})) + C

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What is the integral of e^{3x}cos(x/2) ?
The integral of e^{3x}cos(x/2) is 2(1/37 e^{3x}sin(x/2)+6/37 e^{3x}cos(x/2))+C