What is the integral of zsin(yz) ?

The integral of zsin(yz) is -cos(zy)+C

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integral of zsin(yz)

\int z sin (yz) d y

- cos (zy) + C

Solution steps

\int z sin (yz) d y

Take the constant out:

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

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

Apply u-substitution

= z \cdot \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

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

Use the common integral:

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

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

Substitute back

u = zy

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

Simplify

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

= - cos (zy)

Add a constant to the solution

= - cos (zy) + C