What is the integral of (sin(3x))/3 ?

The integral of (sin(3x))/3 is -1/9 cos(3x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sin(3x))/3

\int \frac{sin ( 3 x )}{3} d x

- \frac{1}{9} cos (3 x) + C

Solution steps

\int \frac{sin ( 3 x )}{3} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int sin (3 x) d x

Apply u-substitution

: \int sin (u) \frac{1}{3} d u

= \frac{1}{3} \cdot \int sin (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{3} \cdot \frac{1}{3} (- cos (u))

Substitute back

u = 3 x

= \frac{1}{3} \cdot \frac{1}{3} (- cos (3 x))

Simplify

\frac{1}{3} \cdot \frac{1}{3} (- cos (3 x)) : - \frac{1}{9} cos (3 x)

= - \frac{1}{9} cos (3 x)

Add a constant to the solution

= - \frac{1}{9} cos (3 x) + C

\int \frac{- x}{1 + x} d x

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

\int_{0}^{2} x d x

f (x) = 1 0^{x}

\int sin (10 t) d t

What is the integral of (sin(3x))/3 ?
The integral of (sin(3x))/3 is -1/9 cos(3x)+C