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

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

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

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

π

Decimal Notation

3.14159 \dots

Solution steps

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

Complete the square

2 x - x^{2} : - (x - 1)^{2} + 1

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

π

= π

\int_{0}^{2} π x d x

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