We can separate the integral.
f(x) = int(cotx)dx - int(tan2x)dx
f(x) = int(cosx/sinx)dx - int((sin2x)/(cos2x))dx
For int(cosx/sinx)dx
Let u = sinx, then du = cosxdx and dx = (du)/cosx.
int(tanx) = int(cosx/u)((du)/cosx) = int(1/u)du = ln|u| = ln|sinx|
For int(tan2x):
Let u = 2x. Then du = 2dx -> dx = 1/2du.
int(sin2x)/(cos2x)dx = int(sinu/cosu)1/2du = 1/2int(sinu/cosu)du
Let v = cosu. Then dv = -sinudu and du = (dv)/(-sinu).
1/2int(sinu)/(cosu)du = 1/2int(sinu)/v * (dv)/(-sinu) = -1/2int(1/v)dv = -1/2ln|cosu| = -1/2ln|cos2x|
Putting this together, we get:
f(x) = ln|sinx| + 1/2ln|cos2x| + C -> We can't forget to add the constant, C.
Our initial values are that when x= pi/3, y = -1. Use this to solve for C:
-1 = ln|sin(pi/3)| + 1/2ln|cos(2(pi/3))| + C
-1 = ln(sqrt(3)/2) + 1/2ln|1/2| + C
C = -1 - 1/2ln(3/4) + 1/2ln(1/2)
C = -1 - 1/2(ln(3/4) + ln1/2)
C = -1 - 1/2(ln(3/8))
This can be approximated to C = -0.51.
Therefore, the function with these initial values is f(x) = ln|sinx| + 1/2ln|cos2x| - 0.51.
Hopefully this helps!