tzk221 Wrote:What's wrong with this proof?
integral tan(x) dx
= integral sin(x) / cos (x) dx
= integral sin(x) sec(x)
= -cos(x) sec(x) - integral -cos(x) sec(x) tan(x) dx
= -1 + integral tan(x) dx
integral tan(x) dx = -1 + integral tan(x) dx
0 = -1
Wow, break up lines and use more parentheses. So you used integration by parts and it looks like you let
u = sec(x)
dv = sin(x)
du = sec(x)tan(x)dx
v = -cos(x)
∫udv = vu - ∫vdu
Keep track of your du, dx and what you're substituting.
From here (the bolded part):
-cos(x) sec(x) - ∫-cos(x)du
= -cos(x)sec(x) + [cos(x)] ∫du
= -cos(x)sec(x) + [cos(x)] ∫sec(x)tan(x)dx
= -cos(x)sec(x) + [cos(x)] (sec(x) + c)