What is the integral of (x^2)/(sqrt(16-9x^2)) ?

The integral of (x^2)/(sqrt(16-9x^2)) is 4/27 (2arcsin(3/4 x)-sin(2arcsin(3/4 x)))+C

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integral of (x^2)/(sqrt(16-9x^2))

\int \frac{x ^{2}}{16 - 9 x ^{2}} d x

\frac{4}{27} (2 arcsin (\frac{3}{4} x) - sin (2 arcsin (\frac{3}{4} x))) + C

Solution steps

\int \frac{x ^{2}}{16 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{16 sin ^{2} ( u )}{27} d u

= \int \frac{16 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{16}{27} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= \frac{16}{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{16}{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{16}{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{16}{27} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{3}{4} x)

= \frac{16}{27} \cdot \frac{1}{2} (arcsin (\frac{3}{4} x) - \frac{1}{2} sin (2 arcsin (\frac{3}{4} x)))

Simplify

\frac{16}{27} \cdot \frac{1}{2} (arcsin (\frac{3}{4} x) - \frac{1}{2} sin (2 arcsin (\frac{3}{4} x))) : \frac{4}{27} (2 arcsin (\frac{3}{4} x) - sin (2 arcsin (\frac{3}{4} x)))

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

Add a constant to the solution

= \frac{4}{27} (2 arcsin (\frac{3}{4} x) - sin (2 arcsin (\frac{3}{4} x))) + C

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What is the integral of (x^2)/(sqrt(16-9x^2)) ?
The integral of (x^2)/(sqrt(16-9x^2)) is 4/27 (2arcsin(3/4 x)-sin(2arcsin(3/4 x)))+C