What is the integral of zcos(yz) ?

The integral of zcos(yz) is sin(zy)+C

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integral of zcos(yz)

\int z cos (yz) d y

sin (zy) + C

Solution steps

\int z cos (yz) d y

Take the constant out:

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

= z \cdot \int cos (zy) d y

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= z \frac{1}{z} sin (u)

Substitute back

u = zy

= z \frac{1}{z} sin (zy)

Simplify

z \frac{1}{z} sin (zy) : sin (zy)

= sin (zy)

Add a constant to the solution

= sin (zy) + C

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