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

The integral of (x^2)/(sqrt(4-9x^2)) is 1/27 (2arcsin(3/2 x)-sin(2arcsin(3/2 x)))+C

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

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

\frac{1}{27} (2 arcsin (\frac{3}{2} x) - sin (2 arcsin (\frac{3}{2} x))) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{4 sin ^{2} ( u )}{27} d u

= \int \frac{4 sin ^{2} ( u )}{27} d u

Take the constant out:

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

= \frac{4}{27} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= \frac{4}{27} \cdot \int \frac{1 - cos ( 2 u )}{2} d u

Take the constant out:

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

= \frac{4}{27} \cdot \frac{1}{2} \cdot \int 1 - cos (2 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{4}{27} \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= \frac{4}{27} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

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

= \frac{4}{27} \cdot \frac{1}{2} (arcsin (\frac{3}{2} x) - \frac{1}{2} sin (2 arcsin (\frac{3}{2} x)))

Simplify

\frac{4}{27} \cdot \frac{1}{2} (arcsin (\frac{3}{2} x) - \frac{1}{2} sin (2 arcsin (\frac{3}{2} x))) : \frac{1}{27} (2 arcsin (\frac{3}{2} x) - sin (2 arcsin (\frac{3}{2} x)))

= \frac{1}{27} (2 arcsin (\frac{3}{2} x) - sin (2 arcsin (\frac{3}{2} x)))

Add a constant to the solution

= \frac{1}{27} (2 arcsin (\frac{3}{2} x) - sin (2 arcsin (\frac{3}{2} x))) + C

\frac{\partial}{\partial x} (x^{2} - y^{2} - z^{2})

f (x) = 2 x^{2} - 1

x \to 0 + lim ((1 + x^{2})^{\frac{1}{x}})

y^{^{''}} + 16 y = cos (4 x) + sin (4 x)

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

What is the integral of (x^2)/(sqrt(4-9x^2)) ?
The integral of (x^2)/(sqrt(4-9x^2)) is 1/27 (2arcsin(3/2 x)-sin(2arcsin(3/2 x)))+C