Question

What is `f(x) = int -cos6x -3tanx+cot(x/3) dx` if `f(pi)=-2`?

Answer 1

`f(x) = int( -cos6x -3tanx+cot(x/3)) dx`

`=> f(x) = -intcos6xdx -3inttanxdx+intcot(x/3) dx`

`=> f(x) = -1/6sin6x -3ln|secx|+3ln|sin(x/3)| +c ...color(red)"(1)"`

,where c= integration constant

Now imposing the given condition `f(pi)=-2`

`-1/6sin(6xxpi) -3ln|secpi|+3ln|sin(pi/3)| +c=-2`

`1/6xx0-3xx0+3xxln(sqrt3/2)+c=-2`

`c=-2-3ln(sqrt3/2)`

Inserting the value of c in `equation color(red)"(1)"` we have

`f(x) = -1/6sin6x -3ln|secx|+3ln|sin(x/3)| -2-3ln(sqrt3/2)`