What is the integral of rsqrt(4-r^2) ?

The integral of rsqrt(4-r^2) is -1/3 (4-r^2)^{3/2}+C

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integral of rsqrt(4-r^2)

\int r 4 - r^{2} d r

- \frac{1}{3} (4 - r^{2})^{\frac{3}{2}} + C

Solution steps

\int r 4 - 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 = 4 - r^{2}

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

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (4 - r^{2})^{\frac{3}{2}} : - \frac{1}{3} (4 - r^{2})^{\frac{3}{2}}

= - \frac{1}{3} (4 - r^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (4 - r^{2})^{\frac{3}{2}} + C

\frac{d x}{d y} = - \frac{4 y ^{2} + 10 x y}{5 y ^{2} + 2 x}

x \to 3 lim (f (x) (g (x)))

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 0

\frac{e ^{x}}{x + 4}

\frac{d}{d x} (x^{3} + x^{2} - 8 x + 7)

What is the integral of rsqrt(4-r^2) ?
The integral of rsqrt(4-r^2) is -1/3 (4-r^2)^{3/2}+C