What is the integral of zsqrt(1-z^2) ?

The integral of zsqrt(1-z^2) is -1/3 (1-z^2)^{3/2}+C

integral of zsqrt(1-z^2)

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

- \frac{1}{3} (1 - z^{2})^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

= \int - u^{2} d u

Take the constant out:

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

= - \int u^{2} d u

Apply the Power Rule

= - \frac{u ^{3}}{3}

Substitute back

u = 1 - z^{2}

= - \frac{( 1 - z ^{2} ) ^{3}}{3}

Simplify

- \frac{( 1 - z ^{2} ) ^{3}}{3} : - \frac{1}{3} (1 - z^{2})^{\frac{3}{2}}

= - \frac{1}{3} (1 - z^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (1 - z^{2})^{\frac{3}{2}} + C

\int \frac{x ^{3} + 3 x ^{2} - 9 x - 2}{x - 2} d x

\int \frac{1 + x}{x} \frac{1}{x} d x

\int ∣ 2 x - 4 ∣ d x

\int \frac{sec ( x )}{2 tan ( x )} d x

\int \frac{x}{x - 14} d x

What is the integral of zsqrt(1-z^2) ?
The integral of zsqrt(1-z^2) is -1/3 (1-z^2)^{3/2}+C