We can start from the given
f(x)=int(x*sin x-cos 4x)*dx also if when f(pi)=-2
Integrating by Parts
int u*dv=uv-int v*du
color(red)(int(x*sin x)dx)
let u=x
let dv=sin x *dx
let v=-cos x
let du=dx
int u*dv=uv-int v*du
int(x)(sin x*dx)=x*(-cos x)-int(-cos x)*dx
int(x*sin x)dx=-x cos x+sin x
Integrating the int cos 4x dx
color(red)(int cos 4x dx)
int cos 4x dx=1/4 int cos 4x*4*dx=1/4*sin 4x
Let us have now the complete integration
f(x)=int(x*sin x-cos 4x)*dx
f(x)=-x cos x+sin x-1/4*sin 4x+C
But f(pi)=-2
f(x)=-x cos x+sin x-1/4*sin 4x+C
f(pi)=-(pi) cos (pi)+sin (pi)-1/4*sin 4(pi)+C
-2=-(pi) (-1)+(0)-1/4*(0)+C
-2=pi+0-0+C
C=-pi-2
Final answer
color(red)(f(x)=-x cos x+sin x-1/4*sin 4x-pi-2)
God bless....I hope the explanation is useful.