What is the integral of sqrt(x/(a-x)) ?

The integral of sqrt(x/(a-x)) is a(arcsin(sqrt(x/a))-1/2 sin(2arcsin(sqrt(x/a))))+C

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integral of sqrt(x/(a-x))

\int \frac{x}{a - x} d x

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

Solution steps

\int \frac{x}{a - x} d x

Simplify

\frac{x}{a - x} : \frac{x}{a - x}

= \int \frac{x}{a - x} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Substitute back

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

Simplify

2 a \frac{1}{2} (arcsin (\frac{1}{a} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{a} x))) : a (arcsin (\frac{x}{a}) - \frac{1}{2} sin (2 arcsin (\frac{x}{a})))

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

Add a constant to the solution

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

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