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

The integral of x^2sqrt(9-4x^2) is 81/64 (arcsin(2/3 x)-1/4 sin(4arcsin(2/3 x)))+C

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

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

\frac{81}{64} (arcsin (\frac{2}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{2}{3} x))) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{81}{8} sin^{2} (u) cos^{2} (u) d u

= \int \frac{81}{8} sin^{2} (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{81}{8} \cdot \int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= \frac{81}{8} \cdot \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= \frac{81}{8} \cdot \frac{1}{8} \cdot \int 1 - cos (4 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

= \frac{81}{8} \cdot \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= \frac{81}{8} \cdot \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

u = arcsin (\frac{2}{3} x)

= \frac{81}{8} \cdot \frac{1}{8} (arcsin (\frac{2}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{2}{3} x)))

Simplify

\frac{81}{8} \cdot \frac{1}{8} (arcsin (\frac{2}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{2}{3} x))) : \frac{81}{64} (arcsin (\frac{2}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{2}{3} x)))

= \frac{81}{64} (arcsin (\frac{2}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{2}{3} x)))

Add a constant to the solution

= \frac{81}{64} (arcsin (\frac{2}{3} x) - \frac{1}{4} sin (4 arcsin (\frac{2}{3} x))) + C

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What is the integral of x^2sqrt(9-4x^2) ?
The integral of x^2sqrt(9-4x^2) is 81/64 (arcsin(2/3 x)-1/4 sin(4arcsin(2/3 x)))+C