A possible way of computing this integral #int_(pi/4)^(pi/2)cscx - sinx dx#, is by simplifying to give; #int_(pi/4)^(pi/2)cosxcotx dx# But a more interesting method is via individually finding #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) #
Hence #int csc x- sinxdx # = # cosx - sinh^-1(cotx) +c #
Hence #int_(pi/4)^(pi/2)cscx - sinx dx# = | # cosx - sinh^-1(cotx) # | from #pi/4# to #pi/2#
So hence by using #sinh^-1(x) = ln( x + (1+x^2)^(1/2)) #, We can simply evaluate to give;
#ln(1+2^(1/2)) -2^(-1/2)#
