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

The integral from 0 to 1/2 of sqrt(x/(1-x)) is pi/4-1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\int_{0}^{\frac{1}{2}} \frac{x}{1 - x} d x

\frac{π}{4} - \frac{1}{2}

Decimal

0.28539 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{x}{1 - x} d x

Simplify

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

= \int_{0}^{\frac{1}{2}} \frac{x}{1 - x} d x

Apply u-substitution

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

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

= - \int_{\frac{1}{2}}^{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_{\frac{1}{2}}^{1} - u^{2} + 1 d u)

Apply Trigonometric Substitution

= - (- 2 \cdot \int_{\frac{π}{4}}^{\frac{π}{2}} cos^{2} (v) d v)

Rewrite using trig identities

= - (- 2 \cdot \int_{\frac{π}{4}}^{\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_{\frac{π}{4}}^{\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_{\frac{π}{4}}^{\frac{π}{2}} 1 d v + \int_{\frac{π}{4}}^{\frac{π}{2}} cos (2 v) d v))

\int_{\frac{π}{4}}^{\frac{π}{2}} 1 d v = \frac{π}{4}

\int_{\frac{π}{4}}^{\frac{π}{2}} cos (2 v) d v = - \frac{1}{2}

= - (- 2 \cdot \frac{1}{2} (\frac{π}{4} - \frac{1}{2}))

Simplify

- (- 2 \cdot \frac{1}{2} (\frac{π}{4} - \frac{1}{2})) : \frac{π}{4} - \frac{1}{2}

= \frac{π}{4} - \frac{1}{2}

What is the integral from 0 to 1/2 of sqrt(x/(1-x)) ?
The integral from 0 to 1/2 of sqrt(x/(1-x)) is pi/4-1/2