Question

Find the constant of integration `c` given that `f'(x)=2cosx-3sinx` and `f(pi/4)=1/sqrt2`? i already solved it and im getting `c=-4/sqrt2` but the answer is `c=-2sqrt2`

`c=-2sqrt2`

Answer 1

Please see below.

Explanation:

Both are same. See the last step.

`-4/sqrt2=-2sqrt2`

As `f'(x)=2cosx-3sinx`

`f(x)=int(2cosx-3sinx)dx`

= `2sinx+3cosx+c`

As `f(pi/4)=1/sqrt2`, we have

`2sin(pi/4)+3cos(pi/4)+c=1/sqrt2`

or `2*1/sqrt2+3*1/sqrt2+c=1/sqrt2`

i.e. `c=-4/sqrt2=-(2xx2)/sqrt2=-2*(sqrt2)^2/sqrt2=-2sqrt2`