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

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

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

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

\frac{π - 2}{4}

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Apply Integration By Parts

: [\frac{2 - u}{u} u - \int - \frac{1}{u ^{\frac{1}{2}} - u + 2} d u]_{1}^{2}

= \frac{1}{2} [\frac{2 - u}{u} u - \int - \frac{1}{u ^{\frac{1}{2}} - u + 2} d u]_{1}^{2}

\int - \frac{1}{u ^{\frac{1}{2}} - u + 2} d u = - 2 arcsin (\frac{1}{2} u)

= \frac{1}{2} [\frac{2 - u}{u} u - (- 2 arcsin (\frac{1}{2} u))]_{1}^{2}

Simplify

= \frac{1}{2} [\frac{2 - u}{u} u + 2 arcsin (\frac{1}{2} u)]_{1}^{2}

Compute the boundaries:

π - 1 - \frac{π}{2}

= \frac{1}{2} (π - 1 - \frac{π}{2})

Simplify

= \frac{π - 2}{4}

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