How do you integrate #int 1/sqrt(e^(2x)+12e^x+45)dx# using trigonometric substitution?

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Answer 1 · 2018-07-14T14:32:11

# 1/(3sqrt5)ln|tan[1/2{arctan((e^x-6)/3)-arctan2}]|+C#.

Let, #I=int1/sqrt(e^(2x)+12e^x+45)dx #.

#e^(2x)+12e^x+45=(e^x+6)^2+3^2#.

So, let us subst. #e^x+6=3tany," so that, "e^xdx=3sec^2ydy#.

#:. dx=(3sec^2ydy)/e^x=(3sec^2ydy)/(3tany-6)=(3sec^2ydy)/(3(tany-2))#.

#:. I=int1/sqrt{(3tany)^2+3^2}*(3sec^2ydy)/(3(tany-2))#,

#=1/3int(secydy)/(tany-2)#,

#:. I=1/3intdy/(siny-2cosy)#.

Now, #siny-2cosy=sqrt5*(1/sqrt5siny-2/sqrt5cosy)#.

#"So, if we set "cosalpha=1/sqrt5," then, "sinalpha=2/sqrt5, and, #

#siny-2cosy=sqrt5(sinycosalpha-cosysinalpha)=sqrt5sin(y-alpha)#.

#:. I=1/3int1/(sqrt5sin(y-alpha))dy#,

#=1/(3sqrt5)intcsc(y-alpha)dy#,

#=1/(3sqrt5)ln|tan((y-alpha)/2)|#,

#rArr I=1/(3sqrt5)ln|tan[1/2{arctan((e^x-6)/3)-arctan2}]|+C#.

Answer 2 · 2018-07-14T19:02:57

# -1/(3sqrt5)ln|{(15t+2)/sqrt5+sqrt(45t^2+12t+1)}|+C#,

where, #t=1/(y-6)#.

Here is a way to solve the Problem without using trigo.

substn.

Let, #I=int1/sqrt(e^(2x)+12e^x+45)dx=int1/sqrt{(e^x+6)^2+3^2)dx.#

Letting #(e^x+6)=y, e^xdx=dy, or, dx=dy/e^x=dy/(y-6)#.

#:. I=int1/{(y-6)sqrt(y^2+9)}dy#.

Next, we take,

#(y-6)=1/t.:. dy=-1/t^2dt." Also, "y=6+1/t.#

#:. I=int1/{1/tsqrt{(6+1/t)^2+9}}(-1/t^2)dt#,

#=-int1/sqrt{(6t+1)^2+9t^2}dt#,

#=-int1/sqrt(45t^2+12t+1)dt#,

#=-int1/sqrt{(3sqrt5t+2/sqrt5)+(1/sqrt5)^2}dt#,

#=-1/(3sqrt5)ln|{(3sqrt5t+2/sqrt5)+sqrt(45t^2+12t+1)}|#,

#=-1/(3sqrt5)ln|{(15t+2)/sqrt5+sqrt(45t^2+12t+1)}|+C#,

where, #t=1/(y-6)#.