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#

1 Answer
Feb 5, 2018

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#