Q76 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\text { The value of } \int \dfrac{d x}{\sqrt{2 x-x^2}} \text { is }$
- A$\sin ^{-1}(2 x-1)+C$
- B$\sin ^{-1}(x+1)+C$
- C$-\sqrt{2 x-x^2}+C$
- Correct answerD$\sin ^{-1}(x-1)+C$
Solution
The integral is $I = \int \dfrac{dx}{\sqrt{2x - x^2}}$.
Complete the square for the expression inside the square root:
$2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = 1 - (x - 1)^2$.
Substituting this into the integral:
$I = \int \dfrac{dx}{\sqrt{1 - (x - 1)^2}}$.
Using the standard integral formula $\int \dfrac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\dfrac{u}{a}\right) + C$, where $u = x - 1$ and $a = 1$:
$I = \sin^{-1}\left(\dfrac{x - 1}{1}\right) + C = \sin^{-1}(x - 1) + C$.
Answer: $\sin ^{-1}(x-1)+C$
Complete the square for the expression inside the square root:
$2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = 1 - (x - 1)^2$.
Substituting this into the integral:
$I = \int \dfrac{dx}{\sqrt{1 - (x - 1)^2}}$.
Using the standard integral formula $\int \dfrac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\dfrac{u}{a}\right) + C$, where $u = x - 1$ and $a = 1$:
$I = \sin^{-1}\left(\dfrac{x - 1}{1}\right) + C = \sin^{-1}(x - 1) + C$.
Answer: $\sin ^{-1}(x-1)+C$
Hence, the correct answer is option: 'D'.