We can separate the integral.
f(x) = int(cotx)dx - int(tan2x)dxf(x)=∫(cotx)dx−∫(tan2x)dx
f(x) = int(cosx/sinx)dx - int((sin2x)/(cos2x))dxf(x)=∫(cosxsinx)dx−∫(sin2xcos2x)dx
For int(cosx/sinx)dx∫(cosxsinx)dx
Let u = sinxu=sinx, then du = cosxdxdu=cosxdx and dx = (du)/cosxdx=ducosx.
int(tanx) = int(cosx/u)((du)/cosx) = int(1/u)du = ln|u| = ln|sinx|∫(tanx)=∫(cosxu)(ducosx)=∫(1u)du=ln|u|=ln|sinx|
For int(tan2x)∫(tan2x):
Let u = 2xu=2x. Then du = 2dx -> dx = 1/2dudu=2dx→dx=12du.
int(sin2x)/(cos2x)dx = int(sinu/cosu)1/2du = 1/2int(sinu/cosu)du∫sin2xcos2xdx=∫(sinucosu)12du=12∫(sinucosu)du
Let v = cosuv=cosu. Then dv = -sinududv=−sinudu and du = (dv)/(-sinu)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|12∫sinucosudu=12∫sinuv⋅dv−sinu=−12∫(1v)dv=−12ln|cosu|=−12ln|cos2x|
Putting this together, we get:
f(x) = ln|sinx| + 1/2ln|cos2x| + C ->f(x)=ln|sinx|+12ln|cos2x|+C→ We can't forget to add the constant, CC.
Our initial values are that when x= pi/3x=π3, y = -1y=−1. Use this to solve for CC:
-1 = ln|sin(pi/3)| + 1/2ln|cos(2(pi/3))| + C−1=ln∣∣sin(π3)∣∣+12ln∣∣cos(2(π3))∣∣+C
-1 = ln(sqrt(3)/2) + 1/2ln|1/2| + C−1=ln(√32)+12ln∣∣∣12∣∣∣+C
C = -1 - 1/2ln(3/4) + 1/2ln(1/2)C=−1−12ln(34)+12ln(12)
C = -1 - 1/2(ln(3/4) + ln1/2)C=−1−12(ln(34)+ln12)
C = -1 - 1/2(ln(3/8))C=−1−12(ln(38))
This can be approximated to C = -0.51C=−0.51.
Therefore, the function with these initial values is f(x) = ln|sinx| + 1/2ln|cos2x| - 0.51f(x)=ln|sinx|+12ln|cos2x|−0.51.
Hopefully this helps!
