Compute the integral `I = int_(0)^(1) ("arc sin x")/(sqrt(1 - x^(2)))dx`

Answer

To compute the integral

\[ I = \int_{0}^{1} \frac{\arcsin x}{\sqrt{1 - x^2}} \, dx, \]

we will use a substitution method.

### Step 1: Substitution
Let

\[ t = \arcsin x. \]

Then, we differentiate both sides:

\[ x = \sin t \quad \text{and} \quad dx = \cos t \, dt. \]

### Step 2: Change the limits of integration
When \( x = 0 \):

\[ t = \arcsin(0) = 0. \]

When \( x = 1 \):

\[ t = \arcsin(1) = \frac{\pi}{2}. \]

So, the new limits of integration are from \( 0 \) to \( \frac{\pi}{2} \).

### Step 3: Substitute in the integral
Now, we can rewrite the integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{t}{\sqrt{1 - \sin^2 t}} \cos t \, dt. \]

Since \( \sqrt{1 - \sin^2 t} = \cos t \), we have:

\[ I = \int_{0}^{\frac{\pi}{2}} t \, dt. \]

### Step 4: Evaluate the integral
Now, we can compute the integral:

\[ I = \int_{0}^{\frac{\pi}{2}} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{(\frac{\pi}{2})^2}{2} - \frac{0^2}{2} = \frac{\pi^2}{8}. \]

### Final Result
Thus, the value of the integral is

\[ I = \frac{\pi^2}{8}. \]

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