Evaluate `int sqrt((a-x)/(a+x))dx`

Answer

To evaluate the integral \( \int \sqrt{\frac{a-x}{a+x}} \, dx \), we can follow these steps:

### Step 1: Rationalize the integrand
We start by rewriting the integrand:
\[ \sqrt{\frac{a-x}{a+x}} = \frac{\sqrt{(a-x)(a-x)}}{\sqrt{(a+x)(a-x)}} = \frac{a-x}{\sqrt{(a+x)(a-x)}} \]
Now, we multiply and divide by \( \sqrt{a-x} \):
\[ \int \sqrt{\frac{a-x}{a+x}} \, dx = \int \frac{\sqrt{a-x} \cdot (a-x)}{\sqrt{(a+x)(a-x)}} \, dx \]

### Step 2: Simplify the expression
Using the identity \( (a-b)(a+b) = a^2 - b^2 \), we can rewrite the denominator:
\[ \sqrt{(a+x)(a-x)} = \sqrt{a^2 - x^2} \]
Thus, we have:
\[ \int \frac{(a-x)^{3/2}}{\sqrt{a^2 - x^2}} \, dx \]

### Step 3: Separate the integral
We can separate the integral into two parts:
\[ \int \frac{a^{3/2} - ax^{1/2}}{\sqrt{a^2 - x^2}} \, dx = a \int \frac{dx}{\sqrt{a^2 - x^2}} - \int \frac{x \, dx}{\sqrt{a^2 - x^2}} \]

### Step 4: Use substitution for the second integral
For the second integral, we can use the substitution \( t = a^2 - x^2 \), which gives us \( dt = -2x \, dx \) or \( x \, dx = -\frac{dt}{2} \). The limits will change accordingly, but for indefinite integrals, we focus on the differential:
\[ -\frac{1}{2} \int \frac{dt}{\sqrt{t}} \]

### Step 5: Integrate the terms
Now we can integrate:
1. The first term \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) \)
2. The second term \( -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \cdot 2\sqrt{t} = -\sqrt{a^2 - x^2} \)

### Step 6: Combine the results
Putting everything together, we have:
\[ a \sin^{-1}\left(\frac{x}{a}\right) + \sqrt{a^2 - x^2} + C \]

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