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 $c=-2sqrt2$

Question

$c=-2sqrt2$

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Answer 1 · 2018-02-05T14:30:03

Please see below.

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. $c=-4/sqrt2=-(2xx2)/sqrt2=-2*(sqrt2)^2/sqrt2=-2sqrt2$