What is the integral of (sin(6x))/6 ?

The integral of (sin(6x))/6 is -1/36 cos(6x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sin(6x))/6

\int \frac{sin ( 6 x )}{6} d x

- \frac{1}{36} cos (6 x) + C

Solution steps

\int \frac{sin ( 6 x )}{6} d x

Take the constant out:

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

= \frac{1}{6} \cdot \int sin (6 x) d x

Apply u-substitution

: \int sin (u) \frac{1}{6} d u

= \frac{1}{6} \cdot \int sin (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \frac{1}{6} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{6} \cdot \frac{1}{6} (- cos (u))

Substitute back

u = 6 x

= \frac{1}{6} \cdot \frac{1}{6} (- cos (6 x))

Simplify

\frac{1}{6} \cdot \frac{1}{6} (- cos (6 x)) : - \frac{1}{36} cos (6 x)

= - \frac{1}{36} cos (6 x)

Add a constant to the solution

= - \frac{1}{36} cos (6 x) + C

\int \frac{1}{7 x ^{2} - 8} d x

\int \frac{1 + x ^{2}}{- 5 x} d x

\int \frac{x + 2}{1 - x ^{4}} d x

\int \frac{1}{- x ^{2} + 2 x + 2} d x

\int \frac{1}{1 - 6 x - 9 x ^{2}} d x

What is the integral of (sin(6x))/6 ?
The integral of (sin(6x))/6 is -1/36 cos(6x)+C