Q89 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
$\int\dfrac{1}{\sqrt{2x - x^2}}dx =$
- Correct answerA$\sin^{-1}(x - 1) + c$
- B$\cos^{-1}(x - 1) + c$
- C$\tan^{-1}(x - 1) + c$
- D$\sin^{-1}x + c$
Solution
The given integral is $I = \int\dfrac{1}{\sqrt{2x - x^2}}dx$
Completing the square for the expression inside the square root:
$2x - x^2 = -(x^2 - 2x) = 1 - (x^2 - 2x + 1) = 1 - (x - 1)^2$
Substituting this into the integral:
$I = \int\dfrac{1}{\sqrt{1 - (x - 1)^2}}dx$
Using the standard integration formula $\int\dfrac{1}{\sqrt{a^2 - x^2}}dx = \sin^{-1}\left(\dfrac{x}{a}\right) + c$:
$I = \sin^{-1}(x - 1) + c$
Answer: $\sin^{-1}(x - 1) + c$
Completing the square for the expression inside the square root:
$2x - x^2 = -(x^2 - 2x) = 1 - (x^2 - 2x + 1) = 1 - (x - 1)^2$
Substituting this into the integral:
$I = \int\dfrac{1}{\sqrt{1 - (x - 1)^2}}dx$
Using the standard integration formula $\int\dfrac{1}{\sqrt{a^2 - x^2}}dx = \sin^{-1}\left(\dfrac{x}{a}\right) + c$:
$I = \sin^{-1}(x - 1) + c$
Answer: $\sin^{-1}(x - 1) + c$
Hence, the correct answer is option: 'A'.