What is the integral of sqrt(y/(c-y)) ?

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

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integral of sqrt(y/(c-y))

\int \frac{y}{c - y} d y

c (arcsin (\frac{y}{c}) - \frac{1}{2} sin (2 arcsin (\frac{y}{c}))) + C

Solution steps

\int \frac{y}{c - y} d y

Simplify

\frac{y}{c - y} : \frac{y}{c - y}

= \int \frac{y}{c - y} d y

Apply u-substitution

: \int \frac{2 u ^{2}}{c - u ^{2}} d u

= \int \frac{2 u ^{2}}{c - u ^{2}} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{c - u ^{2}} d u

Apply Trigonometric Substitution

: \int c sin^{2} (v) d v

= 2 \cdot \int c sin^{2} (v) d v

Take the constant out:

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

= 2 c \cdot \int sin^{2} (v) d v

Rewrite using trig identities

= 2 c \cdot \int \frac{1 - cos ( 2 v )}{2} d v

Take the constant out:

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

= 2 c \frac{1}{2} \cdot \int 1 - cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 c \frac{1}{2} (\int 1 d v - \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 2 c \frac{1}{2} (v - \frac{1}{2} sin (2 v))

Substitute back

= 2 c \frac{1}{2} (arcsin (\frac{1}{c} y) - \frac{1}{2} sin (2 arcsin (\frac{1}{c} y)))

Simplify

2 c \frac{1}{2} (arcsin (\frac{1}{c} y) - \frac{1}{2} sin (2 arcsin (\frac{1}{c} y))) : c (arcsin (\frac{y}{c}) - \frac{1}{2} sin (2 arcsin (\frac{y}{c})))

= c (arcsin (\frac{y}{c}) - \frac{1}{2} sin (2 arcsin (\frac{y}{c})))

Add a constant to the solution

= c (arcsin (\frac{y}{c}) - \frac{1}{2} sin (2 arcsin (\frac{y}{c}))) + C

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