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

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

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integral of 1/(sqrt(2θ-θ^2))

\int \frac{1}{2 θ - θ ^{2}} d θ

arcsin (θ - 1) + C

Solution steps

\int \frac{1}{2 θ - θ ^{2}} d θ

Complete the square

2 θ - θ^{2} : - (θ - 1)^{2} + 1

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

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 = θ - 1

= arcsin (θ - 1)

Add a constant to the solution

= arcsin (θ - 1) + C

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