but you were done from the beginningQ ealuate the integral of tan^2(x) bewteen limits 0 and (pie)/4, use substitutionn where possible.
sorry i dont have word..if anyone could suggest n e thing else would be gr8.
k so tan^2(x)x = sec^2(x) -1
and using cos2a = 2cos^2(x) - 1(method 1)
cos^-2(2x) = 2cos2x + 2 (may have gone wrong here not sure)
or integral of cos^-2(x) = -sin^-1(x) (is this wright?)
n e way.. the answer is;........................1 - (pie)/4.............................
I get using method 1 integral of 2cos2x + 1 = sin2x + x between same limits
= sin(pie/2) + (pie)/2 = 1 + (pie)/2.
i get using method two -(squareroot of 2)-(pie)/4...... i flukily re-arranged this into( -4(squareroot of 2) - (pie)(squaeroot of 2) ) all divided by 4(squareroot of two) which gives the right answer but is obviosly a wrong expansion of -(squareroot of 2)-(pie/4)!!!!!!!!!!!!!!!!
sorry again about notation,, is n e one could put the time in to solve/correct my obsessive nature....that would be great.. if or after the man.utd match pwoa!!cheers
we know the integral of and 1.
no substitution necessary