Originally Posted by

**i_zz_y_ill** Q 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(x)=(1/2)cos2x+(1/2)

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