# Evaluate the integral # I = int_(pi/4)^(pi/2) \ 1/sinx \ dx #?

# int_(pi/4)^(pi/2) \ 1/sinx \ dx = ln (sqrt(2) +1) = 0.88137 # (5dp)

We want to evaluate:

The integral is a standard which can be looked up, to get

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To evaluate the integral ( I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\sin(x)} , dx ), you can use a substitution. Let ( u = \sin(x) ), then ( du = \cos(x) , dx ).

So the integral becomes ( I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{u} , du ).

Integrating ( \frac{1}{u} ) with respect to ( u ) yields ( \ln|u| ), so:

[ I = \ln|\sin(x)| \Bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \ln|\sin\left(\frac{\pi}{2}\right)| - \ln|\sin\left(\frac{\pi}{4}\right)| ]

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

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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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