Evaluate $\int \sqrt{2ax - x^2} dx$. |
$\frac{a}{2} \sqrt{a^2 - (x - a)^2} + \frac{x^2}{2} \sin^{-1} \frac{x - a}{a} + C$ $\frac{x^2 - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{x^2}{2} \sin^{-1} \frac{x - a}{a} + C$ $\frac{x - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1} \frac{x - a}{a} + C$ $\frac{x - a}{4} \sqrt{a^2 - (x - a)^2} - \frac{a^2}{2} \sin^{-1} \frac{x - a}{a} + C$ |
$\frac{x - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1} \frac{x - a}{a} + C$ |
The correct answer is Option (3) → $\frac{x - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1} \frac{x - a}{a} + C$ Let, $I = \int \sqrt{2ax - x^2} dx$ $= \int \sqrt{a^2 - a^2 + 2ax - x^2} \cdot dx$ $= \int \sqrt{a^2 - (x^2 - 2ax + a^2)} \cdot dx$ $= \int \sqrt{a^2 - (x - a)^2} \cdot dx \text{}$ Put $x - a = t \Rightarrow dx = dt$ $= \int \sqrt{a^2 - t^2} \cdot dt$ $= \frac{t}{2} \sqrt{a^2 - t^2} + \frac{a^2}{2} \sin^{-1} \frac{t}{a} + C$ $= \frac{x - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1} \frac{x - a}{a} + C \text{}$ |