What is the integral of r^2sqrt(16-r^2) ?

The integral of r^2sqrt(16-r^2) is 32(arcsin(1/4 r)-1/4 sin(4arcsin(1/4 r)))+C

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integral of r^2sqrt(16-r^2)

\int r^{2} 16 - r^{2} d r

32 (arcsin (\frac{1}{4} r) - \frac{1}{4} sin (4 arcsin (\frac{1}{4} r))) + C

Solution steps

\int r^{2} 16 - r^{2} d r

Apply Trigonometric Substitution

: \int 256 sin^{2} (u) cos^{2} (u) d u

= \int 256 sin^{2} (u) cos^{2} (u) d u

Take the constant out:

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

= 256 \cdot \int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= 256 \cdot \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= 256 \cdot \frac{1}{8} \cdot \int 1 - cos (4 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

= 256 \cdot \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= 256 \cdot \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

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

= 256 \cdot \frac{1}{8} (arcsin (\frac{1}{4} r) - \frac{1}{4} sin (4 arcsin (\frac{1}{4} r)))

Simplify

256 \cdot \frac{1}{8} (arcsin (\frac{1}{4} r) - \frac{1}{4} sin (4 arcsin (\frac{1}{4} r))) : 32 (arcsin (\frac{1}{4} r) - \frac{1}{4} sin (4 arcsin (\frac{1}{4} r)))

= 32 (arcsin (\frac{1}{4} r) - \frac{1}{4} sin (4 arcsin (\frac{1}{4} r)))

Add a constant to the solution

= 32 (arcsin (\frac{1}{4} r) - \frac{1}{4} sin (4 arcsin (\frac{1}{4} r))) + C

\frac{d}{d x} ((x^{2}) (x^{2} + 5)^{2})

\frac{d}{d x} (200 x)

\frac{d}{d x} ((x^{- \frac{1}{2}} + 2 x) (7 - x^{- 1}))

f (x) = - 4 x + 5

\frac{1}{x - 5}

What is the integral of r^2sqrt(16-r^2) ?
The integral of r^2sqrt(16-r^2) is 32(arcsin(1/4 r)-1/4 sin(4arcsin(1/4 r)))+C