IntegrationWatch this thread
Because when you are for example integrating sin2x you get -cos2x/2 because you are dividing by the derivative of the bracket (2x), so I’m asking if I have to divide ln(f(x)) by f’(x)?
The best thing you could do would be to rewrite the integrand as (1/2)(2 cos(2x))/(1 + sin(2x)), and then factorise the 1/2 oustide the integral sign. The integrand is then of the exact form f'(x)/f(x).
Although looking through this again now, if you mean "do I divide by 2?", then it is a yes. But I prefer to get an integrand of the exact form f'(x)/f(x) before integrating, it may take an extra line that doesn't do anything particularly clever, but it makes it very clear what you are doing.
If you have something simple like f(x) = sin(kx) where k is a constant, then it's certainly true that f'(x) = kcos(kx), so if you are integrating cos(kx) then it would be correct to divide by k to get (1/k)sin(kx). But it's dangerous to assume that you can divide by arbitrary derivatives when you're integrating a ratio - it's almost always better to use a substitution until you're more confident
| 1-2sin^2(x)/1+2sin(x)co(x)dx= | cos(2x)/1+sin(2) dx= u=sin(2x)-- apply integration by Substitution change variables to u=sin(2x) implies
implies du/dx=2cos(2x) remodify equation ---- 1
| 1/2du/dx / 1+u dx= 1/2(1+u)du integrate wrt u ----> 1/2In(1+u)= [ 1/2In(1+sin(2x)] x=pi/4 upper, lower=0