What is the integral of (-3x^2)/(sqrt(16-x^2)) ?

The integral of (-3x^2)/(sqrt(16-x^2)) is -24(arcsin(1/4 x)-1/2 sin(2arcsin(1/4 x)))+C

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integral of (-3x^2)/(sqrt(16-x^2))

\int \frac{- 3 x ^{2}}{16 - x ^{2}} d x

- 24 (arcsin (\frac{1}{4} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} x))) + C

Solution steps

\int \frac{- 3 x ^{2}}{16 - x ^{2}} d x

Take the constant out:

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

= - 3 \cdot \int \frac{x ^{2}}{16 - x ^{2}} d x

Apply Trigonometric Substitution

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

= - 3 \cdot \int 16 sin^{2} (u) d u

Take the constant out:

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

= - 3 \cdot 16 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= - 3 \cdot 16 \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

= - 3 \cdot 16 \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

= - 3 \cdot 16 \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)

= - 3 \cdot 16 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{4} x)

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

Simplify

- 3 \cdot 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} x))) : - 24 (arcsin (\frac{1}{4} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} x)))

= - 24 (arcsin (\frac{1}{4} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} x)))

Add a constant to the solution

= - 24 (arcsin (\frac{1}{4} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{4} x))) + C

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What is the integral of (-3x^2)/(sqrt(16-x^2)) ?
The integral of (-3x^2)/(sqrt(16-x^2)) is -24(arcsin(1/4 x)-1/2 sin(2arcsin(1/4 x)))+C