What is the integral of y^3sqrt(9-y^2) ?

The integral of y^3sqrt(9-y^2) is 1/5 (9-y^2)^{5/2}-3(9-y^2)^{3/2}+C

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integral of y^3sqrt(9-y^2)

\int y^{3} 9 - y^{2} d y

\frac{1}{5} (9 - y^{2})^{\frac{5}{2}} - 3 (9 - y^{2})^{\frac{3}{2}} + C

Solution steps

\int y^{3} 9 - y^{2} d y

Apply u-substitution

: \int - (- u^{2} + 9) u^{2} d u

= \int - (- u^{2} + 9) 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} + 9) u^{2} d u

Expand

(- u^{2} + 9) u^{2} : - u^{4} + 9 u^{2}

= - \int - u^{4} + 9 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 9 u^{2} d u)

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

\int 9 u^{2} d u = 3 u^{3}

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

Substitute back

u = 9 - y^{2}

= - - \frac{( 9 - y ^{2} ) ^{5}}{5} + 3 (9 - y^{2})^{3}

Simplify

- - \frac{( 9 - y ^{2} ) ^{5}}{5} + 3 (9 - y^{2})^{3} : \frac{1}{5} (9 - y^{2})^{\frac{5}{2}} - 3 (9 - y^{2})^{\frac{3}{2}}

= \frac{1}{5} (9 - y^{2})^{\frac{5}{2}} - 3 (9 - y^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (9 - y^{2})^{\frac{5}{2}} - 3 (9 - y^{2})^{\frac{3}{2}} + C

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What is the integral of y^3sqrt(9-y^2) ?
The integral of y^3sqrt(9-y^2) is 1/5 (9-y^2)^{5/2}-3(9-y^2)^{3/2}+C