What is the integral of-4sin(2z) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-4sin(2z)

\int - 4 sin (2 z) d z

2 cos (2 z) + C

Solution steps

\int - 4 sin (2 z) d z

Take the constant out:

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

= - 4 \cdot \int sin (2 z) d z

Apply u-substitution

: \int sin (u) \frac{1}{2} d u

= - 4 \cdot \int sin (u) \frac{1}{2} d u

Take the constant out:

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

= - 4 \cdot \frac{1}{2} \cdot \int sin (u) d u

Use the common integral:

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

= - 4 \cdot \frac{1}{2} (- cos (u))

Substitute back

u = 2 z

= - 4 \cdot \frac{1}{2} (- cos (2 z))

Simplify

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

= 2 cos (2 z)

Add a constant to the solution

= 2 cos (2 z) + C

\int 6 x e^{6 x} d x

\int sin^{2} (3 x + 4) cos (3 x + 4) d x

\int 16 - 2 x d x

\int \frac{3}{9 - 25 x ^{2}} d x

\int \frac{e ^{6 x}}{x} d x