C3 Further Integration- Where did I go wrong?

Q10) Where did I go wrong in my working out? The answer at the back of my textbook says it is 1/2Pi - 1.

tan^2(x) is the the lower curve, so your meant to do the integral of 1/2sec^2(x)-tan^2(x)

Well the way I'd do it is different to you..
I'd intergrate the equation y=1/2sec^2x - tan^2x between the limits of 0 and Pi/4 and then times this answer by two to get the area in between the two curves..
I would do it now but I'm off out so if you haven't worked it out later then quote me on here and I'll get round to helping out a bit better
In the meantime try my way? But good luck

Hmm... still didn't get the same answer as the book.

Hmm... still didn't get the same answer as the book.

You made an error on your first line of working -tan^2(x) becomes 1-sec^2(x)

Your mistake is on the second line - the -1 should be a +1.