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!