Here ,
#f(x)=intxsin4xdx+2int tanxdx=I_1+2I_2#
Now,
#I_1=intxsin4xdx#
Using Integration by parts:
#I_1=x*intsin4xdx-int(1*intsin4xdx)dx#
#=>I_1=x*(-cos(4x)/4)-int(-cos(4x)/4)dx#
#=>I_1=-x/4cos4x+1/4intcos4xdx#
#=>I_1=-x/4cos4x+1/4sin(4x)/4+c_1#
#=>I_1=1/16[sin4x-4xcos4x]+c_1#
Again ,#I_2=int tanxdx=ln|secx|+c_2#
So, #f(x)=I_1+2I_2#
#=>F(x)=1/16[sin4x-4xcos4x]+2ln|secx|+c_1+c_2#
#f(x)=1/16[sin4x-4xcos4x]+ln|sec^2x|+C toC=c_1+c_2#
So,
#f(pi/4)=1/16[sinpi-picospi]+ln|sec^2(pi/4)|+C=1#
#=>1/16[0-pi(-1)]+ln|(sqrt2)^2|+C=1#
#=>pi/16+ln2+C=1#
#=>C=1-pi/16-ln2#
Hence ,
#F(x)=1/16[sin4x-4xcos4x]+ln|sec^2x|+1-pi/16-ln2#