What is the integral of sin(yz) ?

The integral of sin(yz) is -1/z cos(zy)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(yz)

\int sin (yz) d y

- \frac{1}{z} cos (zy) + C

Solution steps

\int sin (yz) d y

Apply u-substitution

: \int \frac{sin ( u )}{z} d u

= \int \frac{sin ( u )}{z} d u

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = zy

= \frac{1}{z} (- cos (zy))

Simplify

= - \frac{1}{z} cos (zy)

Add a constant to the solution

= - \frac{1}{z} cos (zy) + C

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

f (x) = - \frac{7}{x ^{8}}

\frac{d}{d x} ((x + y)^{3})

\frac{d}{d x} ((x^{3} - 8)^{\frac{2}{3}})

2 - 6 x