Find the integral: $\displaystyle \int \frac{dx}{\sqrt{2x - x^2}}$
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\sin^{-1}(x - 1) + C$
$\int \frac{dx}{\sqrt{2x - x^2}} = \int \frac{dx}{\sqrt{1 - (x - 1)^2}}$
Put $x - 1 = t$. Then $dx = dt$.
Therefore, $\int \frac{dx}{\sqrt{2x - x^2}} = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1} (t) + C$
$= \sin^{-1} (x - 1) + C$