How do you integrate #int 1/sqrt(e^(2x)+12e^x+37)dx# using trigonometric substitution?

1 Answer
Mar 11, 2018

#sqrt37/37Ln((tan[1/2*arctan((e^x+1)/6)]+6-sqrt37)/(tan[1/2*arctan((e^x+1)/6)]+6+sqrt37))+C#

Explanation:

#int (dx)/sqrt(e^(2x)+12e^x+37)#

=#int (dx)/sqrt((e^x+1)^2+6^2)#

=#int (e^x*dx)/[e^x*sqrt((e^x+1)^2+6^2)]#

After using #e^x+1=6tany#, #e^x=6tany-1# and #e^x*dx=6(secu)^2*du# transforms, this integral became

#int (6(secy)^2*dy)/[(6tany-1)*6secy]#

=#int (secy*dy)/[(6tany-1)#

After using #z=tan(y/2)#, #dy=(2dz)/(z^2+1)#, #tany=(2z)/(1-z^2)# and #secy=(1+z^2)/(1-z^2)# transforms, it became

=#int [(1+z^2)/(1-z^2)(2dz)/(z^2+1)]/[6*(2z)/(1-z^2)-1]#

=#int [(2z*dz)/(1-z^2)]/[(z^2-12z-1)/(1-z^2)]#

=#int (2dz)/(z^2+12z-1)#

=#int (2dz)/[(z+6)^2-37]#

=#int (2dz)/[(z+6+sqrt37)*(z+6-sqrt37)]#

=#sqrt37/37int dz/(z+6-sqrt37)-sqrt37/37int dz/(z+6+sqrt37)#

=#sqrt37/37Ln(z+6-sqrt37)#-#sqrt37/37Ln(z+6+sqrt37)+C#

=#sqrt37/37Ln((z+6-sqrt37)/(z+6+sqrt37))+C#

=#sqrt37/37Ln((tan(y/2)+6-sqrt37)/(tan(y/2)+6+sqrt37))+C#

After using #e^x+1=6tany# and #y=arctan((e^x+1)/6)# inverse transforms, I found

#int (dx)/sqrt(e^(2x)+12e^x+37)#

=#sqrt37/37Ln((tan[1/2*arctan((e^x+1)/6)]+6-sqrt37)/(tan[1/2*arctan((e^x+1)/6)]+6+sqrt37))+C#