What is #f(x) = int cotx-tan2x dx# if #f(pi/3)=-1 #?

1 Answer
Jan 2, 2017

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!