What is the integral of ((x^2))/(sqrt(81-x^2)) ?

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

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integral of ((x^2))/(sqrt(81-x^2))

\int \frac{( x ^{2} )}{81 - x ^{2}} d x

\frac{81}{2} (arcsin (\frac{1}{9} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{9} x))) + C

Solution steps

\int \frac{x ^{2}}{81 - x ^{2}} d x

Apply Trigonometric Substitution

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

= \int 81 sin^{2} (u) d u

Take the constant out:

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

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

Rewrite using trig identities

= 81 \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

= 81 \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

= 81 \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)

= 81 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{9} x)

= 81 \cdot \frac{1}{2} (arcsin (\frac{1}{9} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{9} x)))

Simplify

81 \cdot \frac{1}{2} (arcsin (\frac{1}{9} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{9} x))) : \frac{81}{2} (arcsin (\frac{1}{9} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{9} x)))

= \frac{81}{2} (arcsin (\frac{1}{9} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{9} x)))

Add a constant to the solution

= \frac{81}{2} (arcsin (\frac{1}{9} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{9} x))) + C

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

(2, 6), (6, 6)

\frac{d}{d x} (- 6 sin (x) - 3 x)

\int x^{4} (x^{5} - 7)^{6} d x

\frac{8}{s - 3}

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