What is the integral of z/(sqrt(1-y^2z^2)) ?

The integral of z/(sqrt(1-y^2z^2)) is arcsin(zy)+C

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integral of z/(sqrt(1-y^2z^2))

\int \frac{z}{1 - y ^{2} z ^{2}} d y

arcsin (zy) + C

Solution steps

\int \frac{z}{1 - y ^{2} z ^{2}} d y

Take the constant out:

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

= z \cdot \int \frac{1}{1 - z ^{2} y ^{2}} d y

Apply Trigonometric Substitution

: \int \frac{1}{z} d u

= z \cdot \int \frac{1}{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 1 d u

Integral of a constant:

\int a d x = a x

= z \frac{1}{z} \cdot 1 \cdot u

Substitute back

u = arcsin (zy)

= z \frac{1}{z} \cdot 1 \cdot arcsin (zy)

Simplify

z \frac{1}{z} \cdot 1 \cdot arcsin (zy) : arcsin (zy)

= arcsin (zy)

Add a constant to the solution

= arcsin (zy) + C

y = \frac{x}{7} - \frac{7}{x}

θ \to 0 lim (\frac{s}{d})

x \to 0 lim (x ln (x) - x)

3 + t^{5} + sin (π t)

\frac{\partial}{\partial x} (\frac{y}{x + z})

What is the integral of z/(sqrt(1-y^2z^2)) ?
The integral of z/(sqrt(1-y^2z^2)) is arcsin(zy)+C