Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

Evaluate $\int \sqrt{2ax - x^2} dx$.

Options:

$\frac{a}{2} \sqrt{a^2 - (x - a)^2} + \frac{x^2}{2} \sin^{-1} \frac{x - a}{a} + C$

$\frac{x^2 - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{x^2}{2} \sin^{-1} \frac{x - a}{a} + C$

$\frac{x - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1} \frac{x - a}{a} + C$

$\frac{x - a}{4} \sqrt{a^2 - (x - a)^2} - \frac{a^2}{2} \sin^{-1} \frac{x - a}{a} + C$

Correct Answer:

$\frac{x - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1} \frac{x - a}{a} + C$

Explanation:

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 $x - a = t \Rightarrow dx = dt$

$= \int \sqrt{a^2 - t^2} \cdot dt$

$= \frac{t}{2} \sqrt{a^2 - t^2} + \frac{a^2}{2} \sin^{-1} \frac{t}{a} + C$

$= \frac{x - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1} \frac{x - a}{a} + C \text{}$