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, $int_0^(pi/2) sec2xdx$

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

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

How do you evaluate the integral int sec2xdxint sec2xdx from 0 to pi/2pi/2?