Question

How do you integrate `int 1/sqrt(-e^(2x)+12e^x-27)dx` using trigonometric substitution?

Answer 1

`\color{blue}{2/{3\sqrt3}\tan^{-1}(\frac{2\tan (1/2sin^{-1}(\frac{e^x-6}{3}))+1}{\sqrt3})+C`

Explanation:

`\int \frac{1}{\sqrt{-e^{2x}+12e^x-27}}\ dx`

`=\int \frac{1}{\sqrt{9-(e^{2x}-12e^x+36)}}\ dx`

`=\int \frac{1}{\sqrt{9-(e^{x}-6)^2}}\ dx`

Let `e^x-6=3\sin\theta\implies e^x\ dx=3\cos\theta\ d\theta`

or `dx=\frac{\cos\theta\ d\theta}{\sin\theta+2}`,

`=\int \frac{1}{\sqrt{9-9\sin^2\theta}}\ \frac{\cos\theta\ d\theta}{\sin\theta+2}`

`=\int \frac{1}{3\cos\theta}\ \frac{\cos\theta\ d\theta}{\sin\theta+2}`

`=1/3\int \frac{d\theta}{\sin\theta+2}`

`=1/3\int \frac{d\theta}{\frac{2\tan(\theta/2)}{1+\tan^2(\theta/2)}+2}`

`=1/6\int \frac{(1+\tan^2(\theta/2))\ \ d\theta}{\tan(\theta/2)+1+\tan^2(\theta/2)}`

`=1/3\int \frac{1/2\sec^2(\theta/2)\ \ d\theta}{(\tan(\theta/2)+1/2)^2+3/4}`

`=1/3\int \frac{d(\tan(\theta/2)+1/2)}{(\tan(\theta/2)+1/2)^2+(\sqrt3/2)^2}`

Using standard formula: `\int \frac{dt}{t^2+a^2}=1/a\tan^{-1}(t/a)`,

`=1/3 (1/{\sqrt3/2})\tan^{-1}(\frac{\tan(\theta/2)+1/2}{\sqrt3/2})+C`

`=2/{3\sqrt3}\tan^{-1}(\frac{2\tan(\theta/2)+1}{\sqrt3})+C`

`=2/{3\sqrt3}\tan^{-1}(\frac{2\tan (1/2sin^{-1}(\frac{e^x-6}{3}))+1}{\sqrt3})+C`

Answer 2

`I=2/(3sqrt3)arc tan((e^x-2sqrt(-e^(2x)+12e^x-27))/(sqrt3(e^x-6)))+c`

Explanation:

Here ,

`I=int1/sqrt(-e^(2x)+12e^x-27)dx`

`-e^(2x)+12e^x-27=9-e^(2x)+12e^x-36=9-(e^x-6)^2`

`:.I=int1/sqrt(9-(e^x-6)^2)dx`

Subst. `color(red)(e^x-6=3sinu)=>e^x=6+3sinu=>e^xdx=3cosudu`

`=>dx=(3cosudu)/(6+3sinu) andcolor(red)( sinu=(e^x-6)/3`

So,

`I=int1/sqrt(9-9sin^2u)*(3cosudu)/(6+3sinu)`

`=int1/(3cosu) xx (3cosu du)/(6+3sinu)`

`=int1/(6+3sinu)du=1/3int1/(2+sinu)du`

Subst. `color(blue)(tan(u/2)=t=>sec^2(u/2)*1/2du=dt=>du=(2dt)/(1+tan^2(u/2))`

`=>du=(2dt)/(1+t^2) and sinu=(2t)/(1+t^2)`

`:.I=1/3int1/(2+(2t)/(1+t^2)) xx(2dt)/(1+t^2)`

`=2/3int1/(2+2t^2+2t)dt=1/3int1/(t^2+t+1)dt`

`=1/3int1/(t^2+t+1/4+3/4)dt=1/3int1/((t+1/2)^2+(sqrt3/2)^2)dt`

`=1/3 1/(sqrt3/2)arc tan((t+1/2)/(sqrt3/2))+c`

`:.I=2/(3sqrt3)arctan((2t+1)/sqrt3)+c......to(A)`

Now, `color(blue)(t=tan(u/2)`

#=>t^2=tan^2(u/2)=(1-cosu)/(1+cosu)=(1-cosu)/(1+cosu) xxcolor(green)((1-cosu)/(1-cosu))#

`=>t^2=(1-cosu)^2/(1-cos^2u)=(1-cosu)^2/sin^2u`

`=>t=(1-cosu)/sinu=(1-sqrt(1-sin^2u))/sinu`

Subst. back `color(red)(sinu=(e^x-6)/3`

`:.t=(1-sqrt(1-((e^x-6)/3)^2))/((e^x-6)/3)`

`:.t=(1-sqrt(9-(e^x-6)^2)/3)/((e^x-6)/3)=(3-sqrt(-e^(2x)+12x-27))/(e^x-6)`

`:.2t+1=(6-2sqrt(-e^(2x)+12e^x-27))/(e^x-6)+1`

`:.2t+1=((6-2sqrt(-e^(2x)+12e^x-27)+e^x-6))/(e^x-6)`

`:.2t+1=(e^x-2sqrt(-e^(2x)+12e^x-27))/(e^x-6)`

So, from `(A)`

`I=2/(3sqrt3)arc tan((e^x-2sqrt(-e^(2x)+12e^x-27))/(sqrt3(e^x-6)))+c`