Given f(x)=int(x*sin 2x-sec 4x) dx if f(pi/12)=-2
We do the integration separately
integration by parts for int(x sin 2x) dx
Let u=x and dv=sin 2x dx
Let v=-1/2 cos 2x" "and du=dx
integration by parts formula
int u dv=uv-int v du
int(x sin 2x) dx=x*(-1/2 cos 2x)-int (-1/2 cos 2x) dx
int(x sin 2x) dx=-x/2 cos 2x+1/4 sin 2x+C_1
the other integral int sec 4x dx
use the formula
int sec u *du=ln (sec u+tan u)+C_2
int sec 4x dx=1/4int sec 4x* 4*dx=1/4*ln (sec 4x+tan 4x)+C_2
The combination
f(x)=int(x*sin 2x-sec 4x) dx
f(x)=-x/2 cos 2x+1/4 sin 2x-1/4*ln (sec 4x+tan 4x)+C_0
If f(pi/12)=-2 we can solve for C_0
-2=-pi/12*1/2 cos 2(pi/12)+1/4 *sin 2(pi/12)-1/4*ln (sec 4(pi/12)+tan 4(pi/12))+C_0
-2=-pi/24* sqrt3/2+1/4 *1/2-1/4*ln (2+sqrt3)+C_0
-2=(-pi* sqrt3)/48+1/8-1/4*ln (2+sqrt3)+C_0
C_0=(pi* sqrt3)/48-17/8+1/4*ln (2+sqrt3)
Final answer
color(red)(f(x)=-x/2 cos 2x+1/4 sin 2x-1/4*ln (sec 4x+tan 4x)+(pi* sqrt3)/48-17/8+1/4*ln (2+sqrt3))
God bless...I hope the explanation is useful.