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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Expand

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

= 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

= 2 (- (- \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}

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

Substitute back

u = 9 - y^{2}

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

Simplify

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

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

Add a constant to the solution

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