`int_0^a(dx)/(sqrt(a x-x^2))`

Answer

To solve the integral \( I = \int_0^a \frac{dx}{\sqrt{ax - x^2}} \), we will follow a systematic approach:

### Step 1: Rewrite the integrand
We start with the expression under the square root:
\[ \sqrt{ax - x^2} = \sqrt{- (x^2 - ax)} = \sqrt{-1 \cdot (x^2 - ax)} \]
To make the expression easier to handle, we can factor out a negative sign:
\[ = \sqrt{-(x^2 - ax)} = \sqrt{-(x^2 - ax + \frac{a^2}{4} - \frac{a^2}{4})} = \sqrt{\frac{a^2}{4} - (x - \frac{a}{2})^2} \]
Thus, we can rewrite the integral as:
\[ I = \int_0^a \frac{dx}{\sqrt{\frac{a^2}{4} - (x - \frac{a}{2})^2}} \]

### Step 2: Change of variables
Let \( u = x - \frac{a}{2} \). Then \( dx = du \) and the limits change as follows:
- When \( x = 0 \), \( u = 0 - \frac{a}{2} = -\frac{a}{2} \)
- When \( x = a \), \( u = a - \frac{a}{2} = \frac{a}{2} \)

Now the integral becomes:
\[ I = \int_{-\frac{a}{2}}^{\frac{a}{2}} \frac{du}{\sqrt{\frac{a^2}{4} - u^2}} \]

### Step 3: Recognize the integral form
The integral \( \int \frac{du}{\sqrt{a^2 - u^2}} \) is known to be equal to \( \sin^{-1} \left( \frac{u}{a} \right) + C \). Thus, we can evaluate:
\[ I = \int_{-\frac{a}{2}}^{\frac{a}{2}} \frac{du}{\sqrt{\frac{a^2}{4} - u^2}} = 2 \sin^{-1} \left( \frac{u}{\frac{a}{2}} \right) \text{ evaluated from } -\frac{a}{2} \text{ to } \frac{a}{2} \]

### Step 4: Evaluate the integral
Evaluating at the limits:
\[ = 2 \left[ \sin^{-1} \left( \frac{\frac{a}{2}}{\frac{a}{2}} \right) - \sin^{-1} \left( \frac{-\frac{a}{2}}{\frac{a}{2}} \right) \right] \]
This simplifies to:
\[ = 2 \left[ \sin^{-1}(1) - \sin^{-1}(-1) \right] \]
Since \( \sin^{-1}(1) = \frac{\pi}{2} \) and \( \sin^{-1}(-1) = -\frac{\pi}{2} \):
\[ = 2 \left[ \frac{\pi}{2} - (-\frac{\pi}{2}) \right] = 2 \left[ \frac{\pi}{2} + \frac{\pi}{2} \right] = 2 \pi \]

### Final Answer
Thus, the value of the integral is:
\[ I = \pi \]