What is the integral of r^3sqrt(1-r^2) ?

The integral of r^3sqrt(1-r^2) is -1/15 (3r^2+2)(-r^2+1)^{3/2}+C

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integral of r^3sqrt(1-r^2)

\int r^{3} 1 - r^{2} d r

- \frac{1}{15} (3 r^{2} + 2) (- r^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int r^{3} 1 - r^{2} d r

Apply u-substitution

: \int - \frac{( - u + 1 ) u}{2} d u

= \int - \frac{( - u + 1 ) 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 + 1) u d u

Expand

(- u + 1) u : - u^{\frac{3}{2}} + u

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

= - \frac{1}{2} (- \int u^{\frac{3}{2}} d u + \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= - \frac{1}{2} (- \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - r^{2}

= - \frac{1}{2} (- \frac{2}{5} (1 - r^{2})^{\frac{5}{2}} + \frac{2}{3} (1 - r^{2})^{\frac{3}{2}})

Simplify

- \frac{1}{2} (- \frac{2}{5} (1 - r^{2})^{\frac{5}{2}} + \frac{2}{3} (1 - r^{2})^{\frac{3}{2}}) : - \frac{1}{15} (3 r^{2} + 2) (- r^{2} + 1)^{\frac{3}{2}}

= - \frac{1}{15} (3 r^{2} + 2) (- r^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{15} (3 r^{2} + 2) (- r^{2} + 1)^{\frac{3}{2}} + C

\frac{d}{d x} (3 \frac{3 x}{6 x - 5})

y = x, y = x^{2}, y = 1

\int \frac{cos ( \frac{π}{x ^{3}} )}{x ^{4}} d x

lo g_{4} (sec^{2} (5 x^{3} - 4 x^{2}))

\int \frac{2 x - 4 + 3 x}{x} d x

What is the integral of r^3sqrt(1-r^2) ?
The integral of r^3sqrt(1-r^2) is -1/15 (3r^2+2)(-r^2+1)^{3/2}+C