How do I evaluateĀ #int_(1/6)^(1/2)[csc(pit)][cot(pit)] dt#?

1 Answer
Mar 14, 2015

For any integral, first try straightaway integration, then substitution. Do I know a function whose derivative is or involves csc time cot?

Yes, of course: #d/(dx)(cscx)=-cscxcotx#.
That's not quite what we're integrating, but we can fix that with a substitution:
Let #u=pit#. That makes #du=pi dt# and #dt = (du)/pi#
When #x=1/6# (the lower limit of integration), then #u=4*1/6=pi/6#.
When #x=1/2# (the upper limit). then #u=pi/2#. So, substituting, our integral becomes:

#int_(pi/6)^(pi/2)cscucotu (du)/pi=1/piint_(pi/6)^(pi/2)cscucotu du#

#=1/pi(-cscu)]_(pi/6)^(pi/2)=1/pi(-csc(pi/2)+csc(pi/6))#

#-1/pi(-1+2)=1/pi#