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

1 Answer
Oct 30, 2017

Integral is divergent

Explanation:

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