What is f(x) = int xsin4x+tanx dx if f(pi/4)=1 ?

1 Answer
Feb 15, 2017

1/16(-4xcos(4x)+sin(4x)-16ln cosx-pi+15-8ln 2)

Explanation:

f(x)=int(-1/4xd(cos(4x))-1/((sinx)') d(cosx))

=-1/4(x cos(4x)-intcos(4x dx))-lncosx

=-1/4xcos(4x)+1/16sin(4x)-logcosx +C

As f(pi/4)=1,

pi/16+1/16-ln(1/sqrt2)+C=1

So, C =-pi/16+15/16-1/2ln2. And so,

f(x)=-1/4xcos(4x)+1/16sin(4x)-logcosx -pi/16+15/16-1/2ln2

=1/16(-4xcos(4x)+sin(4x)-16ln cosx-pi+15-8ln 2)