Evaluate: `intxsin^(- 1)[1/2sqrt((2a-x)/a)]dx`

Answer

To evaluate the integral \( I = \int x \sin^{-1} \left( \frac{1}{2} \sqrt{\frac{2a - x}{a}} \right) dx \), we will follow these steps:

### Step 1: Substitution
Let \( x = 2a \cos \theta \). Then, we differentiate both sides:
\[ dx = -2a \sin \theta \, d\theta \]

### Step 2: Substitute in the Integral
Substituting \( x \) and \( dx \) into the integral:
\[ I = \int (2a \cos \theta) \sin^{-1} \left( \frac{1}{2} \sqrt{\frac{2a - 2a \cos \theta}{a}} \right) (-2a \sin \theta) \, d\theta \]

### Step 3: Simplifying the Argument of the Inverse Sine
The argument simplifies as follows:
\[ \frac{1}{2} \sqrt{\frac{2a(1 - \cos \theta)}{a}} = \frac{1}{2} \sqrt{2(1 - \cos \theta)} = \frac{1}{2} \sqrt{2 \cdot 2 \sin^2 \frac{\theta}{2}} = \sin \frac{\theta}{2} \]
Thus, we have:
\[ I = -4a^2 \int \cos \theta \sin \theta \sin^{-1} \left( \sin \frac{\theta}{2} \right) \, d\theta \]

### Step 4: Using Trigonometric Identities
Using the identity \( \sin^{-1}(\sin x) = x \) for \( x \) in the range of the inverse sine function:
\[ I = -4a^2 \int \cos \theta \sin \theta \cdot \frac{\theta}{2} \, d\theta \]

### Step 5: Integration by Parts
Let \( u = \frac{\theta}{2} \) and \( dv = \sin(2\theta) d\theta \). Then, \( du = \frac{1}{2} d\theta \) and \( v = -\frac{1}{2} \cos(2\theta) \):
\[ I = -4a^2 \left( u v - \int v \, du \right) \]

### Step 6: Substitute Back
After performing the integration by parts and simplifying, we substitute back \( \theta \) in terms of \( x \):
\[ \theta = \cos^{-1} \left( \frac{x}{2a} \right) \]

### Step 7: Final Expression
After substituting and simplifying, we arrive at the final result:
\[ I = a^2 \left( \cos^{-1} \left( \frac{x}{2a} \right) \left( x^2 - 2a^2 \right) - \frac{1}{4} \sin(2\theta) + C \right) \]