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#