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

The integral of 1/(sqrt(2x-x^2)) is arcsin(x-1)+C

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integral of 1/(sqrt(2x-x^2))

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

arcsin (x - 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arcsin (u)

Substitute back

u = x - 1

= arcsin (x - 1)

Add a constant to the solution

= arcsin (x - 1) + C

\frac{1}{x + 2}

\int_{0}^{\frac{π}{3}} sec (x) d x

\int \frac{x}{5 x + 1} d x

2 x + 7

(2 x y^{2} - 5) d x + (2 x^{2} y + 4) d y = 0

What is the integral of 1/(sqrt(2x-x^2)) ?
The integral of 1/(sqrt(2x-x^2)) is arcsin(x-1)+C