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Determine whether each integral is convergent or divergent. Evaluate those that are convergent.$$\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$$

the integral is convergent to $\frac{\pi}{2}$

Calculus 2 / BC

Chapter 6

TECHNIQUES OF INTEGRATION

Section 6

Improper Integrals

Integration Techniques

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in this problem were given the improper integral de axe over the square root of one minus X squared from 0 to 1 and were asked to determine whether this in a role is convergent or divergent. So since this integral is discontinuous, X equals one. We can rewrite this as the limit as t approaches one from zero to t of our integral. So I'm going to begin by just solving the indefinite integral D X over one minus x squared. And this is actually the formula for sine inverse. So this is just sign in verse of X plus e, and we can now rewrite this with our bounds. That would be sine inverse of x from zero to t. If we plug these bounds in, we get sine inverse of t minus sine inverse of zero. And finally, we can take the limit of this the limit as T approaches one of sine inverse of tea minus zero, because sine inverse of zero is just zero. And the sine inverse of one is high house, which means that are integral converges to high halves

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