Question

How do you integrate `int 1/sqrt(-e^(2x)+12e^x+72)dx` using trigonometric substitution?

Answer 1

`I`=`1/(6sqrt2)ln|(6sqrt3-sqrt(-e^(2x)+12x+72)+(sqrt3-sqrt2)(e^x-6))/(6sqrt3-sqrt(-e^(2x)+12x+72)+(sqrt3+sqrt2)(e^x-6))|` `+c`

Explanation:

Here,

`I=int1/sqrt(-e^(2x)+12e^x+72)dx`

Now,

#-e^(2x)+12e^x+72=color(brown)(108)-e^(2x)+12e^x- color(brown)(36)=108-(e^x-6)^2#

`:.I=int 1/sqrt(108-(e^x-6)^2)dx`

Substitute ,#color(red)(e^x-6=sqrt108sinu)=>e^x=sqrt108 sinu+6#

`=>e^xdx`=`sqrt108cosudu=>dx=(sqrt108sinu)/(e^x)`=# (sqrt108cosu)/(sqrt108sinu+6)#

So. #I=int1/sqrt(108-108sin^2u)* (sqrt108cosu)/(sqrt108sinu+6)du#

`I=int1/(sqrt108cosu) xx (sqrt108cosu)/(sqrt108sinu+6)du`

`=int1/(sqrt108sinu+6)du`

`=1/6int1/(sqrt3sinu+1)du`

Substitute , `color(blue)(tan(u/2)=t)=>sec^2(u/2)1/2du=dt`

`=>du=(2dt)/(1+tan^2(u/2))=>color(blue)(du=(2dt)/(1+t^2)) and`

`sinu=(2tan(u/2))/(1+tan^2(u/2))=>color(blue)(sinu=(2t)/(1+t^2)`

So , `I=1/6int1/(sqrt3((2t)/(1+t^2))+1)*(2dt)/(1+t^2)`

`=>I=2/6int1/(2sqrt3t+1+t^2)dt`

`=>I=2/6int1/(t^2+2sqrt3t+3-2)dt to`[completing square]

`=>I=1/3int1/((t+sqrt3)^2-(sqrt2)^2)dt`

#=>I=1/3*1/(2sqrt2)int [((t+sqrt3+sqrt2)-(t+sqrt3- sqrt2))/((t+sqrt3+sqrt2)(t+sqrt3-sqrt2))]dt#

`=1/(6sqrt2)int[1/(t+sqrt3-sqrt2)-1/(t+sqrt3+sqrt2)]dt`

`=1/(6sqrt2)[ln|t+sqrt3-sqrt2|-ln|t+sqrt3+sqrt2|]+c`

`:.I=1/(6sqrt2){ln|(t+sqrt3-sqrt2)/(t+sqrt3+sqrt2)|}+c...to(1)`

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 `sinu=(e^x-6)/sqrt108`

`t=(1-sqrt(1-((e^x-6)/sqrt108)^2))/((e^x-6)/sqrt108)=>color(violet)(t=(sqrt108-sqrt(108-(e^x-6)^2))/(e^x-6))`

`:. t=(6sqrt3-sqrt(-e^(2x)+12x+72))/(e^x-6)`

Hence ,from `(1)`

`I=1/(6sqrt2){ln|(color(violet)((6sqrt3-sqrt(-e^(2x)+12x+72))/(e^x-6))+sqrt3-sqrt2)/(color(violet)((6sqrt3-sqrt(-e^(2x)+12x+72))/(e^x-6))+sqrt3+sqrt2)|}+c`

`I`=`1/(6sqrt2)ln|(6sqrt3-sqrt(-e^(2x)+12x+72)+(sqrt3-sqrt2)(e^x-6))/(6sqrt3-sqrt(-e^(2x)+12x+72)+(sqrt3+sqrt2)(e^x-6))|` `+c`

Answer 2

`1/{6\sqrt2}\ln|\frac{e^x(\sqrt3-\sqrt2)-\sqrt{-e^{2x}+12e^x+72}+6\sqrt2}{e^x(\sqrt3+\sqrt2)-\sqrt{-e^{2x}+12e^x+72}-6\sqrt2}|+C`

Explanation:

Given that

`\int \frac{1}{\sqrt{-e^{2x}+12e^x+72}}\ dx`

`=\int \frac{1}{\sqrt{108-((e^{x})^2-2(6)e^x+6^2)}}\ dx`

`=\int \frac{1}{\sqrt{108-(e^x-6)^2}}\ dx`

Let `e^x-6=6\sqrt3\sin \theta\implies e^x\ dx=6\sqrt3\cos\theta\ d\theta`

`\implies dx=\frac{6\sqrt3\cos\theta\ d\theta}{6\sqrt3\sin\theta+6}`

`\therefore \int \frac{dx}{\sqrt{108-(e^x-6)^2}}`

`=\int \frac{\frac{6\sqrt3\cos\theta\ d\theta}{6\sqrt3\sin\theta+6}}{\sqrt{108-108\sin^2\theta}}`

`=\int \frac{6\sqrt3\cos\theta\ d\theta}{6\sqrt3cos\theta(6\sqrt3\sin\theta+6)}`

`=\int \frac{ d\theta}{6\sqrt3\sin\theta+6}`

`=\int \frac{ d\theta}{6\sqrt3(\frac{2\tan(\theta/2)}{1+\tan^2(\theta/2)})+6}`

`=\int \frac{ (1+\tan^2(\theta/2))d\theta}{12\sqrt3\tan(\theta/2)+6(1+\tan^2(\theta/2))}`

`=1/{6}\int \frac{ sec^2(\theta/2)d\theta}{\tan^2(\theta/2)+2\sqrt3\ \tan(\theta/2)+1}`

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

`=1/{3}\int \frac{ d(\tan(\theta/2)+\sqrt3)}{(\tan(\theta/2)+\sqrt3)^2-(\sqrt2)^2}`

Using standard formula: `\int \frac{dt}{t^2-a^2}=1/{2a}\ln|\frac{t-a}{t+a}|`

`=1/{3}\frac{1}{2\sqrt2}\ln|\frac{\tan(\theta/2)+\sqrt3-\sqrt2}{\tan(\theta/2)+\sqrt3+\sqrt2}|+C`

`=1/{6\sqrt2}\ln|\frac{\frac{6\sqrt3-\sqrt{-e^{2x}+12e^x+72}}{e^x-6}+\sqrt3-\sqrt2}{\frac{6\sqrt3-\sqrt{-e^{2x}+12e^x+72}}{e^x-6}+\sqrt3+\sqrt2}|+C`

`=1/{6\sqrt2}\ln|\frac{e^x(\sqrt3-\sqrt2)-\sqrt{-e^{2x}+12e^x+72}+6\sqrt2}{e^x(\sqrt3+\sqrt2)-\sqrt{-e^{2x}+12e^x+72}-6\sqrt2}|+C`