Evaluate the following integrals:
`int sqrt(2ax-x^2dx`

Answer

To evaluate the integral \( I = \int \sqrt{2ax - x^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start with the expression under the square root:
\[ \sqrt{2ax - x^2} \]
We can rearrange this to make it easier to work with:
\[ \sqrt{-(x^2 - 2ax)} = \sqrt{-(x^2 - 2ax + a^2 - a^2)} = \sqrt{-( (x-a)^2 - a^2 )} \]
This can be rewritten as:
\[ \sqrt{a^2 - (x-a)^2} \]

### Step 2: Substitute to simplify the integral
Now, we can set \( u = x - a \), which gives us \( du = dx \) and \( x = u + a \). The integral becomes:
\[ I = \int \sqrt{a^2 - u^2} \, du \]

### Step 3: Use the standard integral formula
The integral \( \int \sqrt{a^2 - u^2} \, du \) is a standard integral, and its result is:
\[ \int \sqrt{a^2 - u^2} \, du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{u}{a}\right) + C \]

### Step 4: Substitute back to original variable
Now we substitute back \( u = x - a \):
\[ I = \frac{x - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x - a}{a}\right) + C \]

### Final Result
Thus, the evaluated integral is:
\[ I = \frac{x - a}{2} \sqrt{a^2 - (x - a)^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x - a}{a}\right) + C \]

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