# Thread: integral of tan^2(x )between o and (pie)/4

1. ## integral of tan^2(x )between o and (pie)/4

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

2. 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
but you were done from the beginning

$\tan^2 x = \sec^2 x - 1$

we know the integral of $\sec^2 x$ and 1.

Hint: $\frac d{dx} \tan x = \sec^2 x$

no substitution necessary