# How do you evaluate the definite integral #int 4/sin^2(z) dz# from #[pi/6, pi/2]#?

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To evaluate the definite integral (\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{4}{\sin^2(z)} , dz), we can use the properties of integrals and trigonometric identities to simplify the expression and then evaluate it.

First, we rewrite the integrand using the identity (\sin^2(z) = 1 - \cos^2(z)).

[\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{4}{\sin^2(z)} , dz = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{4}{1 - \cos^2(z)} , dz]

Now, we can use the substitution (u = \cos(z)), so (du = -\sin(z) , dz).

[\int \frac{-4}{1 - u^2} , du]

This is a standard integral that can be evaluated using partial fraction decomposition. After performing the decomposition and integrating, we obtain:

[\int \frac{-4}{1 - u^2} , du = -2 \ln \left| \frac{1 + u}{1 - u} \right| + C]

Now, we reverse the substitution:

[-2 \ln \left| \frac{1 + \cos(z)}{1 - \cos(z)} \right| + C]

Finally, we evaluate the integral from (\frac{\pi}{6}) to (\frac{\pi}{2}):

[-2 \ln \left| \frac{1 + \cos(\frac{\pi}{2})}{1 - \cos(\frac{\pi}{2})} \right| - (-2 \ln \left| \frac{1 + \cos(\frac{\pi}{6})}{1 - \cos(\frac{\pi}{6})} \right|)]

Simplify the trigonometric terms:

[-2 \ln \left| \frac{1}{0} \right| - (-2 \ln \left| \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}} \right|)]

The natural logarithm of 0 is undefined, so the integral is also undefined.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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