Question

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

Answer 1

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

Explanation:

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

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

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

Explanation:

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)`.