(Original post by **rahaydenuk**)
Sounds fun, you should try it and see if you get a valid answer!

the bizarre thing is that it does give pi/4!!

basically i write I as S[pi/2,0] 1/(tant + 1) dt

then let u = tant => dt/du = 1/(1 + u^2)

=> I = S[inf,0] 1/[(u+1)(1+u^2)] du

now use partial fractions and you end up with (i think)

I = (1/2)S[inf,0] 1/(u+1) + 1/(1+u^2) - u/(1+u^2) du

which hopefully comes out to be

[ 1/4 ln[(u^2+2u+1)/(1+u^2)] + 1/2 arctanu ] {inf,0}

both logs come out as ln 1 ie 0 and you end up with 1/2(pi/2) = pi/4

is this method vaguely correct? Apart from it being long and the improper integral it all seems valid (i hope!)