Question

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

Answer 1

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

Explanation:

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`