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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{cos(4x)}sin(4x)

\int e^{c o s (4 x)} sin (4 x) d x

- \frac{1}{4} e^{c o s (4 x)} + C

Solution steps

\int e^{c o s (4 x)} sin (4 x) d x

Apply u-substitution

: \int e^{c o s (u)} sin (u) \frac{1}{4} d u

= \int e^{c o s (u)} sin (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{c o s (u)} sin (u) d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{4} \cdot \int - e^{v} d v

Take the constant out:

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

= \frac{1}{4} (- \int e^{v} d v)

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{4} (- e^{v})

Substitute back

= \frac{1}{4} (- e^{c o s (4 x)})

Simplify

= - \frac{1}{4} e^{c o s (4 x)}

Add a constant to the solution

= - \frac{1}{4} e^{c o s (4 x)} + C

\frac{5}{r ^{3}}

\frac{d}{d x} (e^{x + 1} - 3 e^{x} + 2 e^{x^{3}})

t^{2} y^{^{''}} + 6 t y^{^{'}} + 6 y = 0

2 x y^{2} + 4 = 2 (3 - x^{2} y) y^{^{'}}

\int \frac{1}{4} cos (2 x) d x

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