`int ( dx)/( sqrt( 2x - x^(2)))` is equal to

Options

A. `sin^(-1) ( x-1) +C`
B. ` sin^(-1) ( x+ 1) + C`
C. ` - sqrt( 2x - x^(2)) + C`
D. none of these

Answer

To solve the integral \( \int \frac{dx}{\sqrt{2x - x^2}} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{\sqrt{2x - x^2}} \]
We can rearrange the expression under the square root:
\[ 2x - x^2 = - (x^2 - 2x) = - (x^2 - 2x + 1 - 1) = - ((x - 1)^2 - 1) \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{dx}{\sqrt{1 - (x - 1)^2}} \]

### Step 2: Use a Trigonometric Substitution
Recognizing the form of the integral, we can use the substitution:
\[ x - 1 = \sin(\theta) \quad \Rightarrow \quad dx = \cos(\theta) d\theta \]
This transforms our integral into:
\[ I = \int \frac{\cos(\theta) d\theta}{\sqrt{1 - \sin^2(\theta)}} \]
Since \( \sqrt{1 - \sin^2(\theta)} = \cos(\theta) \), we have:
\[ I = \int \frac{\cos(\theta) d\theta}{\cos(\theta)} = \int d\theta \]

### Step 3: Integrate
The integral of \( d\theta \) is simply:
\[ I = \theta + C \]

### Step 4: Back Substitute
Now we need to substitute back for \( \theta \):
\[ \theta = \arcsin(x - 1) \]
Thus, we have:
\[ I = \arcsin(x - 1) + C \]

### Final Answer
The final result for the integral is:
\[ \int \frac{dx}{\sqrt{2x - x^2}} = \arcsin(x - 1) + C \]