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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(2x)e^{cos(2x)}

\int sin (2 x) e^{c o s (2 x)} d x

- \frac{1}{2} e^{c o s (2 x)} + C

Solution steps

\int sin (2 x) e^{c o s (2 x)} d x

Apply u-substitution

: \int sin (u) e^{c o s (u)} \frac{1}{2} d u

= \int sin (u) e^{c o s (u)} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (u) e^{c o s (u)} d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{2} \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}{2} (- \int e^{v} d v)

Use the common integral:

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

= \frac{1}{2} (- e^{v})

Substitute back

= \frac{1}{2} (- e^{c o s (2 x)})

Simplify

= - \frac{1}{2} e^{c o s (2 x)}

Add a constant to the solution

= - \frac{1}{2} e^{c o s (2 x)} + C

\int 7 x^{2} + 3 + \frac{6}{x ^{2} + 1} d x

\frac{d}{d x} (\frac{16}{( 5 - x ) ^{2}} - 1)

(\frac{π}{2})^{^{'}}

\int_{0}^{4} 16 - x^{2} d x

3 x^{e} + 2 e^{x}

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