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

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

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

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

\frac{1}{2} (9 arcsin (\frac{x}{3}) - x 9 - x^{2}) + C

Solution steps

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

Apply Trigonometric Substitution

: \int 9 sin^{2} (u) d u

= \int 9 sin^{2} (u) d u

Take the constant out:

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

= 9 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 9 \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

= 9 \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

= 9 \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)

= 9 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{3} x)

= 9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{3} x)))

Simplify

9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{3} x))) : \frac{1}{2} (9 arcsin (\frac{x}{3}) - x 9 - x^{2})

= \frac{1}{2} (9 arcsin (\frac{x}{3}) - x 9 - x^{2})

Add a constant to the solution

= \frac{1}{2} (9 arcsin (\frac{x}{3}) - x 9 - x^{2}) + C

\frac{d}{d x} (ln (sin (3^{x^{2}})))

(x^{2} + y^{2}) d x - 2 x y d y = 0

1 - ln (3 x)

f (x) = 3 x^{2}, at x = 1

\frac{d}{d x} (x (4 x + 1)^{\frac{1}{3}})

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