What is f(x) = int -cos6x -3tanx dx if f(pi)=-1 ?

1 Answer
Sep 8, 2016

Answer is:

f(x)=-1/6sin(6x)+3ln|cosx|-1

Explanation:

f(x)=int(-cos6x-3tanx)dx

f(x)=-intcos(6x)dx-3inttanxdx

For the first integral:

6x=u

(d(6x))/(dx)=(du)/dx

6=(du)/dx

dx=(du)/6

Therefore:

f(x)=-intcosu(du)/6-3intsinx/cosxdx

f(x)=-1/6intcosudu-3int((-cosx)')/cosxdx

f(x)=-1/6intcosudu+3int((cosx)')/cosxdx

f(x)=-1/6sinu+3ln|cosx|+c

f(x)=-1/6sin(6x)+3ln|cosx|+c

Since f(π)=-1

f(π)=-1/6sin(6π)+3ln|cosπ|+c

-1=-1/6*0+3ln|-1|+c

-1=3ln1+c

c=-1

Therefore:

f(x)=-1/6sin(6x)+3ln|cosx|-1