Question

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

Answer 1

`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)`