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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

: \int 4 e^{8 u} cos (u) d u

= \int 4 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

= 4 \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

= 4 (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

= 4 (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

= 4 (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

= 4 (e^{8 u} sin (u) - 8 (- e^{8 u} cos (u) - (- 8 \cdot \int e^{8 u} cos (u) d u)))

Therefore

4 \cdot \int e^{8 u} cos (u) d u = 4 (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

= 4 (\frac{e ^{8 u} sin ( u )}{65} + \frac{8 e ^{8 u} cos ( u )}{65})

Substitute back

u = \frac{x}{4}

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{1}{( y ^{2} - 3 y + 2 )} y^{^{'}} = \frac{1}{x}

x \to 3 lim (x - 10)

x \to - a lim (f (x))

\frac{d y}{d x} + \frac{3 y}{x} = - \frac{1}{x}

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