How do you evaluate the integral #int sec2xdx# from 0 to #pi/2#?

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Answer 1 · 2017-04-26T09:31:26

#int_0^(pi/2) sec2xdx=0#

Let #2x=u# then #2dx=du#

Now #intsec2xdx=1/2intsecudu#

= #1/2int(secu(secu+tanu))/(secu+tanu)du#

= #1/2int(sec^2u+secutanu)/(tanu+secu)du#

As #d/dx(tanu+secu)=sec^2x+secxtanx#, the above is

= #1/2ln|secu+tanu|+c#

= #1/2ln|sec2x+tan2x|+c#

Hence, #int_0^(pi/2) sec2xdx#

= #1/2[ln|sec2x+tan2x|]_0^(pi/2)#

= #1/2[ln|-1+0|-ln|1-0|]=0#