What is the integral from 0 to 1 of sqrt(x/(1-x)) ?

The integral from 0 to 1 of sqrt(x/(1-x)) is pi/2

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integral from 0 to 1 of sqrt(x/(1-x))

\int_{0}^{1} \frac{x}{1 - x} d x

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

\int_{0}^{1} \frac{x}{1 - x} d x

Simplify

\frac{x}{1 - x} : \frac{x}{1 - x}

= \int_{0}^{1} \frac{x}{1 - x} d x

Apply u-substitution

= \int_{1}^{0} - 2 - u^{2} + 1 d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - 2 - u^{2} + 1 d u

Take the constant out:

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

= - (- 2 \cdot \int_{0}^{1} - u^{2} + 1 d u)

Apply Trigonometric Substitution

= - (- 2 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (v) d v)

Rewrite using trig identities

= - (- 2 \cdot \int_{0}^{\frac{π}{2}} \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 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 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 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d v + \int_{0}^{\frac{π}{2}} cos (2 v) d v))

\int_{0}^{\frac{π}{2}} 1 d v = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 v) d v = 0

= - (- 2 \cdot \frac{1}{2} (\frac{π}{2} + 0))

Simplify

- (- 2 \cdot \frac{1}{2} (\frac{π}{2} + 0)) : \frac{π}{2}

= \frac{π}{2}

\int_{e}^{e^{2}} \frac{ln ( x )}{x} d x

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{3}} d x

\int_{0}^{3} x e^{2 x} d x

\int_{0}^{2} - x^{2} + 2 x d x

\int_{0}^{2} (x^{2} - x) d x

What is the integral from 0 to 1 of sqrt(x/(1-x)) ?
The integral from 0 to 1 of sqrt(x/(1-x)) is pi/2