What is the integral from 0 to 3 of x^4(9-x^2)^{3/2} ?

The integral from 0 to 3 of x^4(9-x^2)^{3/2} is (19683pi)/(256)

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integral from 0 to 3 of x^4(9-x^2)^{3/2}

\int_{0}^{3} x^{4} (9 - x^{2})^{\frac{3}{2}} d x

\frac{19683 π}{256}

Decimal Notation

241.54675 \dots

Solution steps

\int_{0}^{3} x^{4} (9 - x^{2})^{\frac{3}{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 6561 sin^{4} (u) cos^{4} (u) d u

= \int_{0}^{\frac{π}{2}} 6561 sin^{4} (u) cos^{4} (u) d u

Take the constant out:

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

= 6561 \cdot \int_{0}^{\frac{π}{2}} sin^{4} (u) cos^{4} (u) d u

Rewrite using trig identities

= 6561 \cdot \int_{0}^{\frac{π}{2}} sin^{4} (u) (1 - sin^{2} (u))^{2} d u

Expand

sin^{4} (u) (1 - sin^{2} (u))^{2} : sin^{4} (u) - 2 sin^{6} (u) + sin^{8} (u)

= 6561 \cdot \int_{0}^{\frac{π}{2}} sin^{4} (u) - 2 sin^{6} (u) + sin^{8} (u) 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

= 6561 (\int_{0}^{\frac{π}{2}} sin^{4} (u) d u - \int_{0}^{\frac{π}{2}} 2 sin^{6} (u) d u + \int_{0}^{\frac{π}{2}} sin^{8} (u) d u)

\int_{0}^{\frac{π}{2}} sin^{4} (u) d u = \frac{3 π}{16}

\int_{0}^{\frac{π}{2}} 2 sin^{6} (u) d u = \frac{5 π}{16}

\int_{0}^{\frac{π}{2}} sin^{8} (u) d u = \frac{35 π}{256}

= 6561 (\frac{3 π}{16} - \frac{5 π}{16} + \frac{35 π}{256})

Simplify

6561 (\frac{3 π}{16} - \frac{5 π}{16} + \frac{35 π}{256}) : \frac{19683 π}{256}

= \frac{19683 π}{256}

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What is the integral from 0 to 3 of x^4(9-x^2)^{3/2} ?
The integral from 0 to 3 of x^4(9-x^2)^{3/2} is (19683pi)/(256)