What is the integral of 8sin(4x) ?

The integral of 8sin(4x) is -2cos(4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 8sin(4x)

\int 8 sin (4 x) d x

- 2 cos (4 x) + C

Solution steps

\int 8 sin (4 x) d x

Take the constant out:

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

= 8 \cdot \int sin (4 x) d x

Apply u-substitution

= 8 \cdot \int sin (u) \frac{1}{4} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int sin (u) d u

Use the common integral:

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

= 8 \cdot \frac{1}{4} (- cos (u))

Substitute back

u = 4 x

= 8 \cdot \frac{1}{4} (- cos (4 x))

Simplify

8 \cdot \frac{1}{4} (- cos (4 x)) : - 2 cos (4 x)

= - 2 cos (4 x)

Add a constant to the solution

= - 2 cos (4 x) + C