#int_(pi/2)^pi 5cscx dx#
This is an improper integral, and it diverges (to #oo#).
(It does not converge.)
Solution:
The integral is improper because #pi# is not in the domain of #csc#, so we need :
#lim_(brarrpi^-)int_(pi/2)^b 5cscx dx=5lim_(brarrpi^-)int_(pi/2)^b cscx dx#
(It's hard to see, but both limits are #brarrpi^-#,)
We'll need to find #intcscxdx#.
#intcscxdx=int1/sinx dx=int1/(2sin(x/2)cos(x/2)) dx#
#=1/2int(sin^2(x/2)+cos^2(x/2))/(sin(x/2)cos(x/2))dx#
#=1/2int(sin(x/2)/cos(x/2)+cos(x/2)/sin(x/2))dx#
#=1/2[-2ln(cos(x/2)+2lnsin(x/2)]+C#
#=lnsin(x/2)-lncos(x/2)+C# #=lntan(x/2)+C#
So the definite integral is:
, #int_(pi/2)^b cscx dx=lntan(b/2)-lntan(pi/4)=lntan(b/2)-ln1=lntan(b/2)#
Finally, as #brarr pi ^-#, we see that #b/2rarr (pi/2)^-# and #tan(b/2)rarroo#, so as #brarr pi ^-#, we get #lntan(b/2)rarroo#
#lim_(brarrpi^-)int_(pi/2)^b 5cscx dx=5lim_(brarrpi^-)int_(pi/2)^b cscx dx=5(oo)=oo#.
The integral diverges.
