CUET Preparation Today
Find $\int \frac{dx}{\sqrt{4x - x^2}}$ |
$\cos^{-1}\left(\frac{x - 2}{2}\right) + C$ $\sin^{-1}\left(\frac{x - 2}{2}\right) + C$ $\frac{1}{2}\sin^{-1}\left(\frac{x - 2}{2}\right) + C$ $\sin^{-1}(x - 2) + C$ |
$\sin^{-1}\left(\frac{x - 2}{2}\right) + C$ |
The correct answer is Option (2) → $\sin^{-1}\left(\frac{x - 2}{2}\right) + C$ Let $I = \int \frac{dx}{\sqrt{4x - x^2}}$ $= \int \frac{dx}{\sqrt{-(x^2 - 4x)}}$ $= \int \frac{dx}{\sqrt{-(x^2 - 4x + 2^2 - 2^2)}}$ $= \int \frac{dx}{\sqrt{-((x-2)^2 - 2^2)}}$ $= \int \frac{dx}{\sqrt{2^2 - (x - 2)^2}}$ $= \sin^{-1} \left( \frac{x - 2}{2} \right) + C$ [∵$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left( \frac{x}{a} \right) + C$] |
