Question

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

Answer 1

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`