What is f(x) = int e^xcosx-secxtan^3x+sinx dx if f(pi/6) = 1 ?

1 Answer
Nov 8, 2017

f(x)=(e^x)/2*(sinx+cosx)+secx-(secx)^3/3-cosx+1+(7sqrt3)/54-1/2*e^(pi/6)*(sqrt3+1)/2

Explanation:

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