What is the integral of zcos(xz) ?

The integral of zcos(xz) is sin(zx)+C

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integral of zcos(xz)

\int z cos (x z) d x

sin (z x) + C

Solution steps

\int z cos (x z) d x

Take the constant out:

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

= z \cdot \int cos (z x) d x

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 = z x

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

Simplify

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

= sin (z x)

Add a constant to the solution

= sin (z x) + C

(x - 4)^{^{'}}

f (x) = \frac{3}{x - 2}

\frac{\partial}{\partial x} (x^{3} + y^{4})

\frac{d}{d x} (- sin (a x) a)

\frac{d y}{d x} + 3 x e^{y} = 0