What is the integral of rsqrt(16-r^2) ?

The integral of rsqrt(16-r^2) is -1/3 (16-r^2)^{3/2}+C

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integral of rsqrt(16-r^2)

\int r 16 - r^{2} d r

- \frac{1}{3} (16 - r^{2})^{\frac{3}{2}} + C

Solution steps

\int r 16 - r^{2} d r

Apply u-substitution

: \int - \frac{u}{2} d u

= \int - \frac{u}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

u = 16 - r^{2}

= - \frac{1}{2} \cdot \frac{2}{3} (16 - r^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (16 - r^{2})^{\frac{3}{2}} : - \frac{1}{3} (16 - r^{2})^{\frac{3}{2}}

= - \frac{1}{3} (16 - r^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (16 - r^{2})^{\frac{3}{2}} + C

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What is the integral of rsqrt(16-r^2) ?
The integral of rsqrt(16-r^2) is -1/3 (16-r^2)^{3/2}+C