If `int(1)/(sqrt(2ax-x^(2)))dx= fog (x)+C` , then

Options

A. `f(x)=sin^(-1)x, and g (x) = (x+a)/(a)`
B. `f(x)=sin^(-1) x ,and g (x) = (x-a)/(a)`
C. `f(x)=cos^(-1)x , and g (x) = (x-a)/(a)`
D. `f(x) =tan^(-1)x and g (x)=(x-a)/(a)`

Explanation

To solve the integral \( \int \frac{1}{\sqrt{2ax - x^2}} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Denominator**:
We start with the expression under the square root:
\[ \sqrt{2ax - x^2} = \sqrt{-(x^2 - 2ax)} = \sqrt{-(x^2 - 2ax + a^2 - a^2)} = \sqrt{-( (x - a)^2 - a^2 )} \]
Thus, we can express it as:
\[ \sqrt{a^2 - (x - a)^2} \]

2. **Substitute in the Integral**:
Now, we substitute this back into the integral:
\[ \int \frac{1}{\sqrt{2ax - x^2}} \, dx = \int \frac{1}{\sqrt{a^2 - (x - a)^2}} \, dx \]

3. **Use a Trigonometric Substitution**:
We can use the substitution \( x - a = a \sin(\theta) \), which implies \( dx = a \cos(\theta) \, d\theta \). The limits of integration will change accordingly, but since we are looking for an indefinite integral, we focus on the substitution:
\[ \sqrt{a^2 - (x - a)^2} = \sqrt{a^2 - a^2 \sin^2(\theta)} = a \cos(\theta) \]

4. **Rewrite the Integral**:
The integral now becomes:
\[ \int \frac{a \cos(\theta)}{a \cos(\theta)} \, d\theta = \int d\theta \]

5. **Integrate**:
The integral of \( d\theta \) is simply:
\[ \theta + C \]

6. **Back Substitute**:
We need to convert back to \( x \). Since \( \theta = \sin^{-1}\left(\frac{x - a}{a}\right) \), we have:
\[ \int \frac{1}{\sqrt{2ax - x^2}} \, dx = \sin^{-1}\left(\frac{x - a}{a}\right) + C \]

7. **Identify \( f(g(x)) \)**:
From the expression \( \sin^{-1}\left(\frac{x - a}{a}\right) \), we can identify:
- \( f(x) = \sin^{-1}(x) \)
- \( g(x) = \frac{x - a}{a} \)

### Final Answer:
Thus, we have:
\[ \int \frac{1}{\sqrt{2ax - x^2}} \, dx = f(g(x)) + C \]
where \( f(x) = \sin^{-1}(x) \) and \( g(x) = \frac{x - a}{a} \).