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