What is the integral of (r^3)/(sqrt(4-r^2)) ?

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

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integral of (r^3)/(sqrt(4-r^2))

\int \frac{r ^{3}}{4 - r ^{2}} d r

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

Solution steps

\int \frac{r ^{3}}{4 - r ^{2}} d r

Apply u-substitution

= \int u^{2} - 4 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{2} d u - \int 4 d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 4 d u = 4 u

= \frac{u ^{3}}{3} - 4 u

Substitute back

u = 4 - r^{2}

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

Simplify

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

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

Add a constant to the solution

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

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