What is the integral of sqrt(4-r^2) ?

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

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integral of sqrt(4-r^2)

\int 4 - r^{2} d r

2 arcsin (\frac{1}{2} r) + sin (2 arcsin (\frac{1}{2} r)) + C

Solution steps

\int 4 - r^{2} d r

Apply Trigonometric Substitution

: \int 4 cos^{2} (u) d u

= \int 4 cos^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

= 4 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{2} r)

= 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} r) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} r)))

Simplify

4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} r) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} r))) : 2 arcsin (\frac{1}{2} r) + sin (2 arcsin (\frac{1}{2} r))

= 2 arcsin (\frac{1}{2} r) + sin (2 arcsin (\frac{1}{2} r))

Add a constant to the solution

= 2 arcsin (\frac{1}{2} r) + sin (2 arcsin (\frac{1}{2} r)) + C

h \to 0 lim (\frac{( h - 3 ) ^{3} + 1}{h})

f (x) = lo g_{2} (1 - 3 x)

\int 7 e^{- 7 x} d x

\frac{d}{d x} (9 x sin (x))

\frac{d}{d x} (\frac{4}{3} π^{3})

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