What is the integral of v^3(a^2-v^2)^{1/2} ?

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

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integral of v^3(a^2-v^2)^{1/2}

\int v^{3} (a^{2} - v^{2})^{\frac{1}{2}} d v

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

Solution steps

\int v^{3} (a^{2} - v^{2})^{\frac{1}{2}} d v

Apply u-substitution

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

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

Expand

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

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

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

\int a^{2} u^{\frac{1}{2}} d u = a^{2} \frac{2}{3} u^{\frac{3}{2}}

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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