What is the integral of x^5sqrt(9-4x^2) ?

The integral of x^5sqrt(9-4x^2) is -1/280 (10x^4+18x^2+27)(-4x^2+9)^{3/2}+C

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integral of x^5sqrt(9-4x^2)

\int x^{5} 9 - 4 x^{2} d x

- \frac{1}{280} (10 x^{4} + 18 x^{2} + 27) (- 4 x^{2} + 9)^{\frac{3}{2}} + C

Solution steps

\int x^{5} 9 - 4 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{2187}{64} sin^{5} (u) cos^{2} (u) d u

= \int \frac{2187}{64} sin^{5} (u) cos^{2} (u) d u

Take the constant out:

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

= \frac{2187}{64} \cdot \int sin^{5} (u) cos^{2} (u) d u

Rewrite using trig identities

= \frac{2187}{64} \cdot \int (1 - cos^{2} (u))^{2} sin (u) cos^{2} (u) d u

Apply u-substitution

: \int - v^{2} (1 - v^{2})^{2} d v

= \frac{2187}{64} \cdot \int - v^{2} (1 - v^{2})^{2} d v

Expand

- v^{2} (1 - v^{2})^{2} : - v^{2} + 2 v^{4} - v^{6}

= \frac{2187}{64} \cdot \int - v^{2} + 2 v^{4} - v^{6} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{2187}{64} (- \int v^{2} d v + \int 2 v^{4} d v - \int v^{6} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{2187}{64} (- \frac{v ^{3}}{3} + \frac{2 v ^{5}}{5} - \frac{v ^{7}}{7})

Substitute back

= \frac{2187}{64} (- \frac{cos ^{3} ( arcsin ( \frac{2}{3} x ) )}{3} + \frac{2 cos ^{5} ( arcsin ( \frac{2}{3} x ) )}{5} - \frac{cos ^{7} ( arcsin ( \frac{2}{3} x ) )}{7})

Simplify

\frac{2187}{64} (- \frac{cos ^{3} ( arcsin ( \frac{2}{3} x ) )}{3} + \frac{2 cos ^{5} ( arcsin ( \frac{2}{3} x ) )}{5} - \frac{cos ^{7} ( arcsin ( \frac{2}{3} x ) )}{7}) : - \frac{1}{280} (10 x^{4} + 18 x^{2} + 27) (- 4 x^{2} + 9)^{\frac{3}{2}}

= - \frac{1}{280} (10 x^{4} + 18 x^{2} + 27) (- 4 x^{2} + 9)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{280} (10 x^{4} + 18 x^{2} + 27) (- 4 x^{2} + 9)^{\frac{3}{2}} + C

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What is the integral of x^5sqrt(9-4x^2) ?
The integral of x^5sqrt(9-4x^2) is -1/280 (10x^4+18x^2+27)(-4x^2+9)^{3/2}+C