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

Answer 1

Integral is divergent

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

#intcscx dx#
let #u# = #cotx# then #du = -csc^2 x dx#
Via quotient rule; #cotx# = # cosx/sinx# hence #d/dx(cotx)# = #((sinx)(-sinx)-(cosx)(cosx))/sin^2 x# Hence = #-csc^2 x #
Hence #(-du)/cscx = cscx dx#
Hence #intcscx dx# becomes #-int (du)/cscx #
Considering # 1 + cot^2 x = csc^2 x #
Hence if #u# = #cotx# then #cscx# = #(1+u^2)^(1/2)#
Hence #-int (du)/cscx # becomes #-int (du)/(1+u^2)^(1/2) #
Then make a new substitution of #u# = #sinhtheta# Hence #du# = #coshtheta d theta#
Now by cosindering #cosh^2 theta - sinh^2 theta = 1 #
#-int (du)/(1+u^2)^(1/2) # becomes #-int d theta#
= #-theta + c#
As #u# = #sinh theta# then #theta# = #arcsinh(u) #
Hence #-arcsinh (u) + c#
Hence as #u# = #cottheta#
Hence #int csc xdx # = #c - sinh^-1(cotx) #
So hence we evaluate the antiderivative from #pi/2# to #pi#, but there is issue in this as # sinh^-1( cot(pi) ) # is undifined
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Answer 2

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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Answer 3

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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Answer from HIX Tutor

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.

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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