You are not logged in.
Pages: 1
intergrate tanh(2x+1)dx
please help!!!
Offline
x=1 and i dont know were the d comes from
Offline
∫ tanh(2x+1) dx = (1/2)ln(cosh(2x+1))
tanh(u) = -i*tan(iu) ( i = √(-1) )
u = 2x+1
du = 2 dx
thus:
(1/2) du = dx
∫ -i*tan(iu) (1/2)du = (-i/2)∫tan(iu) du = (1/2)(ln(|cos(iu)|) + C
cosh(u) = cos(iu)
(1/2) ln(cosh u) + C = (1/2)ln(cosh(2x+1))
There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.
Offline
UHHHHHHHH...OHHHHHHHHHHHH
You are using complex integration, bossk!!!
X'(y-Xβ)=0
Offline
Did I do something wrong? I was kind of just winging it, and my answer matched The Integrator's.
There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.
Offline
It's so much fun messing around with the integrator!
Offline
You could also just say that since
therefore
No need to mess around.
Offline
I always use the integrator when I'm helping someone out, because with integration I tend to make stupid little mistakes, so I always make sure I'm right... so I don't mislead the person asking the question.
There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.
Offline
Pages: 1