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Find each of the following integrals:∫ 1√9 − x2 dx

Question

Find each of the following integrals:∫ 1√9 − x2 dx

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Solution

The integral you're asking about is a standard form of inverse trigonometric integral. Here's how to solve it:

∫ 1/√(9 - x²) dx

This is a standard form of integral that is recognized as:

∫ dx/√(a² - x²) = arcsin(x/a) + C

where a is a constant and C is the constant of integration.

In your case, a = √9 = 3. So, the integral becomes:

= arcsin(x/3) + C

So, ∫ 1/√(9 - x²) dx = arcsin(x/3) + C.

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