# How do you find the integral of #int cscx dx# from pi/2 to pi?

Integral is divergent

This is an interesting intergal, we can find its antiderivative via considering a valid substitution.

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To find the integral of (\int_{\frac{\pi}{2}}^{\pi} \csc(x) , dx), first, recall that (\csc(x) = \frac{1}{\sin(x)}). Using this substitution, the integral becomes:

(\int_{\frac{\pi}{2}}^{\pi} \frac{1}{\sin(x)} , dx)

Now, make the substitution (u = \cos(x)) and (du = -\sin(x) , dx), which gives:

(\int_{0}^{-1} -\frac{1}{u} , du)

After integrating, you'll get:

(-\ln|u| \bigg|_{0}^{-1})

Finally, evaluate at the limits:

(-\ln|-1| - (-\ln|0|))

Since (\ln|0|) is undefined, the integral is divergent.

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To find the integral of ( \int_{\frac{\pi}{2}}^{\pi} \csc(x) , dx ), we first rewrite ( \csc(x) ) as ( \frac{1}{\sin(x)} ). Then, we integrate ( \frac{1}{\sin(x)} ) with respect to ( x ) from ( \frac{\pi}{2} ) to ( \pi ). This integral is equal to ( \ln|\csc(x) + \cot(x)| ), evaluated from ( \frac{\pi}{2} ) to ( \pi ). Substituting ( \pi ) and ( \frac{\pi}{2} ) into the antiderivative and subtracting the result at ( \frac{\pi}{2} ) from the result at ( \pi ) gives the final answer. The integral evaluates to ( \ln|(-1) + 0| - \ln|\infty + 1| ), which simplifies to ( -\ln(1) - \ln(\infty) ). Since ( \ln(1) = 0 ) and ( \ln(\infty) = \infty ), the final result is ( -\infty ).

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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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