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The problem asks to compute the integral:

\[
\int_{-1}^{0} \frac{dx}{\sqrt{1 - x^2}}
\]

or determine if it diverges.

This is a standard integral involving the inverse sine function. The general formula for this type of integral is:

\[
\int \frac{dx}{\sqrt{1 - x^2}} = \arcsin(x) + C
\]

Now, we can apply the limits of integration to this formula:

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

We know that:

\[
\arcsin(0) = 0
\]
and
\[
\arcsin(-1) = -\frac{\pi}{2}
\]

Thus, the result of the integral is:

\[
0 - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2}
\]

So, the correct answer is:

\[
\frac{\pi}{2}
\]

Let me know if you want more details or have any questions! Here are five related questions:

1. How do you derive the arcsine function from basic integrals?
2. Can this integral be solved using a different substitution method?
3. How do we interpret definite integrals involving inverse trigonometric functions geometrically?
4. What is the relationship between arcsine and cosine integrals?
5. What are other standard forms of integrals for inverse trigonometric functions?

**Tip:** When integrating functions involving square roots like \(\sqrt{1 - x^2}\), always check if they match the form of inverse trigonometric integrals!

Вычислите интеграл \(\int_{-1}^{0} \frac{dx}{\sqrt{1 - x^2}}\) или установите его расходимость.

Learn how to evaluate the definite integral \(\int_{-1}^{0} \frac{dx}{\sqrt{1 - x^2}}\) using the arcsine function. This problem demonstrates how to solve integrals involving inverse trigonometric functions.

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