What is the integral of sin(3t-3x) ?

The integral of sin(3t-3x) is 1/3 cos(3t-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(3t-3x)

\int sin (3 t - 3 x) d x

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

Solution steps

\int sin (3 t - 3 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 3 t - 3 x

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

Simplify

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

Add a constant to the solution

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

y = \frac{1}{5 + 3 x}, (1, \frac{1}{8})

y = 5 x - 3

y = 7 + 5 x^{2} - 2 x^{3}

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

\int (x^{3} e^{x^{4}}) d x

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