Question

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

Answer 1

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

Explanation:

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`