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.
