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

The integral of-2e^{-3cos(3x)}sin(3x) is -2/9 e^{-3cos(3x)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int - 2 e^{- 3 c o s (3 x)} sin (3 x) d x

- \frac{2}{9} e^{- 3 c o s (3 x)} + C

Solution steps

\int - 2 e^{- 3 c o s (3 x)} sin (3 x) d x

Take the constant out:

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

= - 2 \cdot \int e^{- 3 c o s (3 x)} sin (3 x) d x

Apply u-substitution

: \int \frac{e ^{u}}{9} d u

= - 2 \cdot \int \frac{e ^{u}}{9} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{9} \cdot \int e^{u} d u

Use the common integral:

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

= - 2 \cdot \frac{1}{9} e^{u}

Substitute back

u = - 3 cos (3 x)

= - 2 \cdot \frac{1}{9} e^{- 3 c o s (3 x)}

Simplify

- 2 \cdot \frac{1}{9} e^{- 3 c o s (3 x)} : - \frac{2}{9} e^{- 3 c o s (3 x)}

= - \frac{2}{9} e^{- 3 c o s (3 x)}

Add a constant to the solution

= - \frac{2}{9} e^{- 3 c o s (3 x)} + C

\frac{\partial}{\partial x} (e^{p (x)} (p (x) + p (x) p (x)))

\int e^{a x + b} d x

\frac{d}{d t} (7 cos^{3} (t))

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

e^{e^{- x}}

What is the integral of-2e^{-3cos(3x)}sin(3x) ?
The integral of-2e^{-3cos(3x)}sin(3x) is -2/9 e^{-3cos(3x)}+C