Thread: [SOLVED] What am I doing wrong?

This has been bugging me for a long time now... Can anyone please help me?
Integrate dx/(sin(x) + cos(x))




I used the identy that t= tan(x/2)
So following the sin and cos substitutaion I have
dx = 2dt/1+t^2
sin(x) = 2t/(1+t^2)
cos(x) = (1-t^2)/(1+t^2)
thus we have
Integral (2dt/(1+t^2))/((2t + 1-t^2)/(1+t^2)
this simplifies to
Integral 2dt/(2t+1-t^2)
then factor down into
Integral -2dt/(t-1)^2 -2
t=sqrt(2)sec(a) +1
Integral -2dt/2(sec(a)^2 -1)
Integral -dt/(tan^2(a))

Am I way off?? there seems to be way to many substitutions or something... I would appreciate it very much!

I try to avoid the half angle formula.
Instead I would multiply by the conjugate on top and bottom.
It works, but you need to break up the fraction into two pieces
and then there's a little partial fraction problem to solve, but it should work.

Wish i would have noticed that!! Thanks! I will try it that way now.

working the other way, this now becomes the answer

Integral (sinx - cosx)dx / (sin^2(x) - cos^2(x))
Integral (sinx - cosx)dx/(2sin^2(x) - 1)
u = sinx
du = cosxdx
Integral (-udu)/(2u^2 - 1)
-(1/4)ln(1-2u^2)
-(1/4)ln(1-2sinx^2)

Originally Posted by codex123

Wish i would have noticed that!! Thanks! I will try it that way now.

I reviewed Salas/Hille a couple of years ago and they too left out that trick. I mentioned it, but they never used my suggestion. What's weird was that 30 years earlier I used that book as an undergraduate and now they pay me to review the same book.

Originally Posted by codex123

working the other way, this now becomes the answer

Integral (sinx - cosx)dx / (sin^2(x) - cos^2(x))
Integral (sinx - cosx)dx/(2sin^2(x) - 1)
u = sinx
du = cosxdx
Integral (-udu)/(2u^2 - 1)
-(1/4)ln(1-2u^2)
-(1/4)ln(1-2sinx^2)

Be careful, your numerator in (-udu)/(2u^2 - 1)
Looks wrong.