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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= r^{2} \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

= r^{2} \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

= r^{2} \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)

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

Substitute back

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

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

Simplify

r^{2} \frac{1}{2} (arcsin (\frac{1}{r} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{r} y))) : \frac{1}{2} r^{2} (arcsin (\frac{y}{r}) + \frac{1}{2} sin (2 arcsin (\frac{y}{r})))

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

Add a constant to the solution

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

y^{^{'}} - \frac{5}{3} y = 1 - \frac{1}{5} t

y y^{^{'}} = x - 1

\frac{d y}{d x} = 25 + y^{2}

\frac{d y}{d x} = 19 x + y - 19

\frac{d}{d s} (ln (s))

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