What is the integral of x^3sqrt(16-x^2) ?

The integral of x^3sqrt(16-x^2) is 1/5 (16-x^2)^{5/2}-16/3 (16-x^2)^{3/2}+C

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integral of x^3sqrt(16-x^2)

\int x^{3} 16 - x^{2} d x

\frac{1}{5} (16 - x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 - x^{2})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 16 - x^{2} d x

Apply u-substitution

: \int - (- u^{2} + 16) u^{2} d u

= \int - (- u^{2} + 16) u^{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int (- u^{2} + 16) u^{2} d u

Expand

(- u^{2} + 16) u^{2} : - u^{4} + 16 u^{2}

= - \int - u^{4} + 16 u^{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

= - (- \int u^{4} d u + \int 16 u^{2} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int 16 u^{2} d u = \frac{16 u ^{3}}{3}

= - (- \frac{u ^{5}}{5} + \frac{16 u ^{3}}{3})

Substitute back

u = 16 - x^{2}

= - (- \frac{( 16 - x ^{2} ) ^{5}}{5} + \frac{16 ( 16 - x ^{2} ) ^{3}}{3})

Simplify

- (- \frac{( 16 - x ^{2} ) ^{5}}{5} + \frac{16 ( 16 - x ^{2} ) ^{3}}{3}) : \frac{1}{5} (16 - x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 - x^{2})^{\frac{3}{2}}

= \frac{1}{5} (16 - x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (16 - x^{2})^{\frac{5}{2}} - \frac{16}{3} (16 - x^{2})^{\frac{3}{2}} + C

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What is the integral of x^3sqrt(16-x^2) ?
The integral of x^3sqrt(16-x^2) is 1/5 (16-x^2)^{5/2}-16/3 (16-x^2)^{3/2}+C