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

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

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

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

π

Decimal

3.14159 \dots

Solution steps

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

Complete the square

x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{1}{4}

= \int_{0}^{1} \frac{1}{- ( x - \frac{1}{2} ) ^{2} + \frac{1}{4}} d x

Apply u-substitution

= \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{2}{- 4 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}}^{\frac{1}{2}} \frac{1}{- 4 u ^{2} + 1} d u

Apply Trigonometric Substitution

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

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} [1 \cdot v]_{- \frac{π}{2}}^{\frac{π}{2}}

Simplify

2 \cdot \frac{1}{2} [1 \cdot v]_{- \frac{π}{2}}^{\frac{π}{2}} : [v]_{- \frac{π}{2}}^{\frac{π}{2}}

= [v]_{- \frac{π}{2}}^{\frac{π}{2}}

Compute the boundaries:

π

= π

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