What is the integral of 1/3 x^3sqrt(9-x^2) ?

The integral of 1/3 x^3sqrt(9-x^2) is 1/15 (9-x^2)^{5/2}-(9-x^2)^{3/2}+C

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integral of 1/3 x^3sqrt(9-x^2)

\int \frac{1}{3} x^{3} 9 - x^{2} d x

\frac{1}{15} (9 - x^{2})^{\frac{5}{2}} - (9 - x^{2})^{\frac{3}{2}} + C

Solution steps

\int \frac{1}{3} 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

= \frac{1}{3} \cdot \int x^{3} 9 - x^{2} d x

Apply u-substitution

= \frac{1}{3} \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

= \frac{1}{3} (- \int (- u^{2} + 9) u^{2} d u)

Expand

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

= \frac{1}{3} (- \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

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

Substitute back

u = 9 - x^{2}

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

Simplify

\frac{1}{3} (- (- \frac{( 9 - x ^{2} ) ^{5}}{5} + 3 (9 - x^{2})^{3})) : \frac{1}{15} (9 - x^{2})^{\frac{5}{2}} - (9 - x^{2})^{\frac{3}{2}}

= \frac{1}{15} (9 - x^{2})^{\frac{5}{2}} - (9 - x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (9 - x^{2})^{\frac{5}{2}} - (9 - x^{2})^{\frac{3}{2}} + C

\frac{d}{d x} f (x) = \frac{12}{( 3 x - 4 ) ^{5}}

\int \frac{- 2}{z ^{3}} d z

y^{^{''}} + 4 y^{^{'}} + 5 y = 15 x + e^{- x}

\frac{d}{d x} (\frac{sin ( x )}{1 + sin ( x )})

\int_{1}^{\infty} (\frac{1}{x ^{3}}) d x

What is the integral of 1/3 x^3sqrt(9-x^2) ?
The integral of 1/3 x^3sqrt(9-x^2) is 1/15 (9-x^2)^{5/2}-(9-x^2)^{3/2}+C