i have a problem in integration
cos^4A+sin^4A
$ ----------------------- dA = ln [1+tanA/1-tanA] + [1/2] sinAcosA
cos^2A - sin^2A
i solved it in this way. can it be done in another way
cos^4A+sin^4A = 1 - 2cos^2A sin^2A = 1 - [1/2] sin^2 [2A]
and cos^2A - sin^2A = cos2A
i took sin2A as t and 2cos2A as dt and simplified
but i couldent get the required answer
is the method correct or will i have to solve the problem in different way
could you elaborate on how you integrated using sin2A as your substitution.
cos2A is in the denominator.
proceeding by the method you have used we get
$ [(1-sin2 2a/2)/cos 2a ]da
this can easily be integrated because on simplifying the numerator we get
$ {(1+cos22a)/2cos2a}da
on splitting the integrand we get
1/2$sec2ada +1/2$cos2ada
which can easily be found
he
i got the same answer
1/2$sec2a** +1/2$cos2a da
but the answer given as
ln [ 1 - tanA / 1 + tanA ] + [1/2]sinA.cosA
the answer given is correct because sec2a-tan2a can easily be simplified to tan(pi/4-a) which is (1-tana )/1+tana
if you think about it the simplification becomes obvious.