What is the integral of x^3sqrt(b^2-x^2) ?

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

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integral of x^3sqrt(b^2-x^2)

\int x^{3} b^{2} - x^{2} d x

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

Solution steps

\int x^{3} b^{2} - x^{2} d x

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = b^{2} - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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