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

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Answer 1 · 2017-02-15T02:36:01

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

#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)#