Question

If `f'(x)=2cosx+sec2x` and `f(pi/3)=2`, find `f(x)`?

Answer 1

`f(x)=2sinx+1/2ln|(tan2x+sec2x)|+2-sqrt3+1/2ln(2+sqrt3)`

Explanation:

As `f'(x)=2cosx+sec2x`

`f(x)=intf'(x)dx`

= `int(2cosx+sec2x)dx`

= `2sinx+1/2ln|(tan2x+sec2x)|+c`

But `f(pi/3)=2`, hence

`2sin(pi/3)+1/2[ln|(tan((2pi)/3)+sec((2pi)/3)|]+c=2`

or `sqrt3+1/2(ln|-sqrt3-2|+c=2`

or `c=2-sqrt3+1/2ln(2+sqrt3)`

Hence `f(x)=2sinx+1/2ln|(tan2x+sec2x)|+2-sqrt3+1/2ln(2+sqrt3)`