How do you integrate #int 1/sqrt(-e^(2x)-12e^x-35)dx# using trigonometric substitution?

Question

1387 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-13T10:19:27

#I=2/sqrt35 tan^-1(sqrt(7(-e^x-5))/(sqrt(5(e^x+7))))+c#

Here,

#I=int 1/sqrt(-e^(2x)-12e^x-35)dx#

Now,

#-e^(2x)-12e^x-35=1-(e^(2x)+12e^x+36)=1-(e^x+6)^2#

#:.I=int1/sqrt(1-(e^x+6)^2)dx#

Subst. #e^x+6=cosu=>e^x=cosu-6=>e^xdx=-sinudu#

#=>dx=-sinu/e^xdu=-sinu/(cosu-6)du=sinu/(6-cosu)du#

#I=int1/sqrt(1-cos^2u)xxsinu/(6-cosu)du#

#=int1/sinuxxsinu/(6-cosu)du#

#=int1/(6-cosu)du#

Again subst. #tan(u/2)=t=>sec^2(u/2)*1/2du=dt#

#=>du=(2dt)/(1+tan^2(u/2))=(2dt)/(1+t^2)and cosu=(1- t^2)/(1+t^2)#
So,#I=int1/(6-(1-t^2)/(1+t^2))xx(2/(1+t^2))dt#

#=int2/(6+6t^2-1+t^2)dt#

#=int2/(7t^2+5)dt#

#=2/7int1/(t^2+(sqrt(5/7))^2)dt#

#=2/7xx1/(sqrt(5/7)) tan^-1(t/(sqrt(5/7)))+c#

#=2/sqrt35tan^-1((sqrt(7) t)/sqrt5)+c ,where, t=tan(u/2)#

#:.I=2/sqrt35tan^-1((sqrt(7) tan(u/2))/sqrt5)+c #

Now,

#tan^2(u/2)=(1-cosu)/(1+cosu),where, cosu=e^x+6#

#:.tan^2(u/2)=(1-(e^x+6))/(1+(e^x+6))=(-e^x-5)/(e^x+7) <0 ..??#

#=>tan(u/2)=sqrt((-e^x-5)/(e^x+7)#
Thus,

#I=2/sqrt35 tan^-1(sqrt(7(-e^x-5))/(sqrt(5(e^x+7))))+c#