What is the integral of 6x^3sqrt(9-x^2) ?

The integral of 6x^3sqrt(9-x^2) is -6(-1/5 (9-x^2)^{5/2}+3(9-x^2)^{3/2})+C

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integral of 6x^3sqrt(9-x^2)

\int 6 x^{3} 9 - x^{2} d x

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

Solution steps

\int 6 x^{3} 9 - x^{2} d x

Take the constant out:

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

= 6 \cdot \int x^{3} 9 - x^{2} d x

Apply u-substitution

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

= 6 \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

= 6 (- \int (- u^{2} + 9) u^{2} d u)

Expand

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

= 6 (- \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

= 6 (- (- \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}

= 6 (- (- \frac{u ^{5}}{5} + 3 u^{3}))

Substitute back

u = 9 - x^{2}

= 6 (- (- \frac{( 9 - x ^{2} ) ^{5}}{5} + 3 (9 - x^{2})^{3}))

Simplify

6 (- (- \frac{( 9 - x ^{2} ) ^{5}}{5} + 3 (9 - x^{2})^{3})) : - 6 (- \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} + 3 (9 - x^{2})^{\frac{3}{2}})

= - 6 (- \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} + 3 (9 - x^{2})^{\frac{3}{2}})

Add a constant to the solution

= - 6 (- \frac{1}{5} (9 - x^{2})^{\frac{5}{2}} + 3 (9 - x^{2})^{\frac{3}{2}}) + C

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What is the integral of 6x^3sqrt(9-x^2) ?
The integral of 6x^3sqrt(9-x^2) is -6(-1/5 (9-x^2)^{5/2}+3(9-x^2)^{3/2})+C