What is the integral of 2cos(x/9) ?

The integral of 2cos(x/9) is 18sin(x/9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2cos(x/9)

\int 2 cos (\frac{x}{9}) d x

18 sin (\frac{x}{9}) + C

Solution steps

\int 2 cos (\frac{x}{9}) d x

Take the constant out:

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

= 2 \cdot \int cos (\frac{x}{9}) d x

Apply u-substitution

: \int cos (u) \cdot 9 d u

= 2 \cdot \int cos (u) \cdot 9 d u

Take the constant out:

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

= 2 \cdot 9 \cdot \int cos (u) d u

Use the common integral:

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

= 2 \cdot 9 sin (u)

Substitute back

u = \frac{x}{9}

= 2 \cdot 9 sin (\frac{x}{9})

Simplify

= 18 sin (\frac{x}{9})

Add a constant to the solution

= 18 sin (\frac{x}{9}) + C

\int cos (\frac{1}{5} t) d t

\int 4 x^{2} + 8 x + 5 d x

\int (\frac{5}{x})^{2} d x

\int (x^{2} + 2 x) ln (x) d x

\int \frac{3 x - 2}{x - 1} d x