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} (\frac{( 2 x ^{3} + a x + a )}{x ^{2} - 1})

\int 4 x^{2} e^{x^{3}} d x

y = - 2 x^{3} + \frac{6}{x}, at x = - 1

\int \frac{x ^{3}}{2 x ^{4} + x ^{2} - 1} d x

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

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