Question #33f91

1 Answer
Mar 6, 2017

int_0^(pi/4)secx^2tanxdx=1/2

Explanation:

Perhaps you mean int_0^(pi/4)secx^2tanxdx

To find this assume u=tanx, then du=sec^2xdx and

Observe that as tan0=0 and tan(pi/4)=1, the new limits are 0 and 1 and

int_0^(pi/4)secx^2tanxdx=int_0^1udu

= [u^2/2]_0^1

= 1/2-0=1/2