f(x)=int e^x*cosx*dx-int secx*(tanx)^3*dx+int sinx*dx
I initially calculated 3 subintegrals,
A=int e^x*cosx*dx
=e^x*sinx-int e^x*sinx*dx
=e^x*sinx-[e^x*(-cosx)-int e^x*(-cosx)*dx]
=e^x*sinx-[-e^x*cosx+int e^x*cosx*dx]
=e^x*sinx+e^x*cosx-int e^x*cosx*dx
=e^x*(sinx+cosx)-A
Hence,
2A=e^x*(sinx+cosx) or A=(e^x)/2*(sinx+cosx)
B=int secx*(tanx)^3*dx
=int (tanx)^2*(secx*tanx)*dx
=int [(secx)^2-1]*(secx*tanx)*dx
=int [(secx)^2*(secx*tanx)*dx-int secx*tanx*dx
=(secx)^3/3-secx
C=int sinx*dx=-cosx+C
So,
f(x)=A-B+C
=(e^x)/2*(sinx+cosx)+secx-(secx)^3/3-cosx+C
After imposing f(pi/6)=1 condition,
1/2*e^(pi/6)*(sqrt3+1)/2+(2sqrt3)/3-(8sqrt(3))/27-sqrt(3)/2+C=1
1/2*e^(pi/6)*(sqrt3+1)/2-(7sqrt(3))/54+C=1
C=1+(7sqrt3)/54-1/2*e^(pi/6)*(sqrt3+1)/2
Thus,
f(x)=(e^x)/2*(sinx+cosx)+secx-(secx)^3/3-cosx+1+(7sqrt3)/54-1/2*e^(pi/6)*(sqrt3+1)/2