Q66 / 180 · COMEDK 2026 (9 May Shift 1) · Mathematics
$\int \sqrt{2ax - x^2}\ dx =$
- Correct answerA$\dfrac{x-a}{2}\sqrt{2ax-x^2} + \dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x-a}{a}\right) + C$
- B$\dfrac{x-a}{2}\sqrt{2ax-x^2} + \dfrac{a^2}{2}\cos^{-1}\left(\dfrac{x-a}{a}\right) + C$
- C$\dfrac{a^2}{2}\sqrt{2ax-x^2} + \dfrac{x-a}{2}\sin^{-1}\left(\dfrac{x-a}{a}\right) + C$
- D$\dfrac{a^2}{2}\sqrt{2ax-x^2} + \dfrac{x-a}{2}\cos^{-1}\left(\dfrac{x-a}{a}\right) + C$
Solution
$I = \int \sqrt{2ax - x^2}\ dx$
Completing the square inside the square root:
$2ax - x^2 = -(x^2 - 2ax) = -(x^2 - 2ax + a^2 - a^2) = a^2 - (x - a)^2$
Substituting this back into the integral:
$I = \int \sqrt{a^2 - (x - a)^2}\ dx$
Using the standard integral formula $\int \sqrt{A^2 - X^2}\ dX = \dfrac{X}{2}\sqrt{A^2 - X^2} + \dfrac{A^2}{2}\sin^{-1}\left(\dfrac{X}{A}\right) + C$
Substituting $X = x - a$ and $A = a$:
$I = \dfrac{x - a}{2}\sqrt{a^2 - (x - a)^2} + \dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x - a}{a}\right) + C$
$I = \dfrac{x - a}{2}\sqrt{2ax - x^2} + \dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x - a}{a}\right) + C$
Answer: $\dfrac{x-a}{2}\sqrt{2ax-x^2} + \dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x-a}{a}\right) + C$
Completing the square inside the square root:
$2ax - x^2 = -(x^2 - 2ax) = -(x^2 - 2ax + a^2 - a^2) = a^2 - (x - a)^2$
Substituting this back into the integral:
$I = \int \sqrt{a^2 - (x - a)^2}\ dx$
Using the standard integral formula $\int \sqrt{A^2 - X^2}\ dX = \dfrac{X}{2}\sqrt{A^2 - X^2} + \dfrac{A^2}{2}\sin^{-1}\left(\dfrac{X}{A}\right) + C$
Substituting $X = x - a$ and $A = a$:
$I = \dfrac{x - a}{2}\sqrt{a^2 - (x - a)^2} + \dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x - a}{a}\right) + C$
$I = \dfrac{x - a}{2}\sqrt{2ax - x^2} + \dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x - a}{a}\right) + C$
Answer: $\dfrac{x-a}{2}\sqrt{2ax-x^2} + \dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x-a}{a}\right) + C$
Hence, the correct answer is option: 'A'.