# How do you evaluate #int cos(z)/sin^7(z) dz# for [pi/2, pi/6]?

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To evaluate the integral of ( \frac{\cos(z)}{\sin^7(z)} ) over the interval ( \left[\frac{\pi}{2}, \frac{\pi}{6}\right] ), you can use the substitution method. Let ( u = \sin(z) ), then ( du = \cos(z) dz ). The integral becomes ( \int \frac{1}{u^7} du ). Integrating this expression yields ( -\frac{1}{6u^6} + C ), where ( C ) is the constant of integration. Now, resubstitute ( u = \sin(z) ) and evaluate the integral from ( \frac{\pi}{2} ) to ( \frac{\pi}{6} ).

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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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