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

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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

= \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}

= - \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

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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