CUET Preparation Today
Find the integral: $\displaystyle \int \frac{dx}{\sqrt{2x - x^2}}$ |
$\sin^{-1}(x - 1) + C$ $\cos^{-1}(x - 1) + C$ $\sin^{-1}(x + 1) + C$ $\log |x-1|\sin^{-1}(x + 1) + C$ |
$\sin^{-1}(x - 1) + C$ |
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, $\int \frac{dx}{\sqrt{2x - x^2}} = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1} (t) + C$ $= \sin^{-1} (x - 1) + C$ |
