Evaluate: `int_sqrt((3a^2+b^2)/2)^sqrt((a^2+b^2)/2) (x*dx)/(sqrt((x^2-a^2)(b^2-x^2)))`

Answer

To evaluate the integral

\[ I = \int_{\sqrt{\frac{3a^2 + b^2}{2}}}^{\sqrt{\frac{a^2 + b^2}{2}}} \frac{x \, dx}{\sqrt{(x^2 - a^2)(b^2 - x^2)}} \]

we will use a substitution method to simplify the integral.

### Step 1: Substitution
Let \( t^2 = b^2 - x^2 \). Then, differentiating both sides gives:

\[ -2x \, dx = 2t \, dt \implies x \, dx = -t \, dt \]

### Step 2: Change of Limits
When \( x = \sqrt{\frac{3a^2 + b^2}{2}} \):

\[ t^2 = b^2 - \left(\frac{3a^2 + b^2}{2}\right) = \frac{2b^2 - 3a^2 - b^2}{2} = \frac{b^2 - 3a^2}{2} \implies t = \sqrt{\frac{b^2 - 3a^2}{2}} \]

When \( x = \sqrt{\frac{a^2 + b^2}{2}} \):

\[ t^2 = b^2 - \left(\frac{a^2 + b^2}{2}\right) = \frac{2b^2 - a^2 - b^2}{2} = \frac{b^2 - a^2}{2} \implies t = \sqrt{\frac{b^2 - a^2}{2}} \]

### Step 3: Rewrite the Integral
Now substituting \( x \, dx \) and changing the limits, we have:

\[ I = -\int_{\sqrt{\frac{b^2 - 3a^2}{2}}}^{\sqrt{\frac{b^2 - a^2}{2}}} \frac{t \, dt}{\sqrt{(b^2 - t^2 - a^2)(t^2)}} \]

### Step 4: Simplifying the Integral
The term \( b^2 - t^2 - a^2 \) can be rewritten as:

\[ b^2 - t^2 - a^2 = b^2 - a^2 - t^2 \]

Thus, the integral becomes:

\[ I = -\int_{\sqrt{\frac{b^2 - 3a^2}{2}}}^{\sqrt{\frac{b^2 - a^2}{2}}} \frac{t \, dt}{\sqrt{(b^2 - a^2 - t^2)(t^2)}} \]

### Step 5: Recognizing the Integral Form
We recognize that the integral

\[ \int \frac{dt}{\sqrt{(A^2 - t^2)(t^2)}} \]

has a known solution, which is related to the inverse sine function.

### Step 6: Evaluating the Integral
Using the known integral result:

\[ \int \frac{dt}{\sqrt{(A^2 - t^2)(t^2)}} = \sin^{-1}\left(\frac{t}{A}\right) + C \]

we can evaluate our integral:

\[ I = -\left[\sin^{-1}\left(\frac{t}{\sqrt{b^2 - a^2}}\right)\right]_{\sqrt{\frac{b^2 - 3a^2}{2}}}^{\sqrt{\frac{b^2 - a^2}{2}}} \]

### Step 7: Substituting Limits
Substituting the limits into the inverse sine function gives:

\[ I = -\left(\sin^{-1}\left(\frac{\sqrt{\frac{b^2 - a^2}{2}}}{\sqrt{b^2 - a^2}}\right) - \sin^{-1}\left(\frac{\sqrt{\frac{b^2 - 3a^2}{2}}}{\sqrt{b^2 - a^2}}\right)\right) \]

### Step 8: Final Result
This simplifies to:

\[ I = -\left(\frac{\pi}{4} - \sin^{-1}\left(\frac{\sqrt{\frac{b^2 - 3a^2}{2}}}{\sqrt{b^2 - a^2}}\right)\right) \]

Thus, the final answer is:

\[ I = \sin^{-1}\left(\frac{\sqrt{\frac{b^2 - 3a^2}{2}}}{\sqrt{b^2 - a^2}}\right) - \frac{\pi}{4} \]