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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int 3 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

= 3 \cdot \int e^{6 u} cos (u) d u

Apply Integration By Parts

= 3 (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

= 3 (e^{6 u} sin (u) - 6 \cdot \int e^{6 u} sin (u) d u)

Apply Integration By Parts

= 3 (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

= 3 (e^{6 u} sin (u) - 6 (- e^{6 u} cos (u) - (- 6 \cdot \int e^{6 u} cos (u) d u)))

Therefore

3 \cdot \int e^{6 u} cos (u) d u = 3 (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

= 3 (\frac{e ^{6 u} sin ( u )}{37} + \frac{6 e ^{6 u} cos ( u )}{37})

Substitute back

u = \frac{x}{3}

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

Simplify

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

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

Add a constant to the solution

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

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