What is the integral of z/(\sqrt[4]{a^2-z^2)} ?

The integral of z/(\sqrt[4]{a^2-z^2)} is -2/3 (-z^2+a^2)^{3/4}+C

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integral of z/(\sqrt[4]{a^2-z^2)}

- \frac{2}{3} (- z^{2} + a^{2})^{\frac{3}{4}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{4}{3} u^{\frac{3}{4}}

= - \frac{1}{2} \cdot \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = a^{2} - z^{2}

= - \frac{1}{2} \cdot \frac{4}{3} (a^{2} - z^{2})^{\frac{3}{4}}

Simplify

- \frac{1}{2} \cdot \frac{4}{3} (a^{2} - z^{2})^{\frac{3}{4}} : - \frac{2}{3} (- z^{2} + a^{2})^{\frac{3}{4}}

= - \frac{2}{3} (- z^{2} + a^{2})^{\frac{3}{4}}

Add a constant to the solution

= - \frac{2}{3} (- z^{2} + a^{2})^{\frac{3}{4}} + C

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What is the integral of z/(\sqrt[4]{a^2-z^2)} ?
The integral of z/(\sqrt[4]{a^2-z^2)} is -2/3 (-z^2+a^2)^{3/4}+C