What is the integral of-9cos(3t) ?

The integral of-9cos(3t) is -3sin(3t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of-9cos(3t)

\int - 9 cos (3 t) d t

- 3 sin (3 t) + C

Solution steps

\int - 9 cos (3 t) d t

Take the constant out:

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

= - 9 \cdot \int cos (3 t) d t

Apply u-substitution

= - 9 \cdot \int cos (u) \frac{1}{3} d u

Take the constant out:

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

= - 9 \cdot \frac{1}{3} \cdot \int cos (u) d u

Use the common integral:

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

= - 9 \cdot \frac{1}{3} sin (u)

Substitute back

u = 3 t

= - 9 \cdot \frac{1}{3} sin (3 t)

Simplify

- 9 \cdot \frac{1}{3} sin (3 t) : - 3 sin (3 t)

= - 3 sin (3 t)

Add a constant to the solution

= - 3 sin (3 t) + C