Question

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

Answer 1

need double check. I'm not absolute sure..but I hope this helps

Explanation:

First of all we need to rearrange the expresion under square root.

Lets do `z=e^x`. With this we have

`sqrt(-z^2+12z-45)=sqrt(-(z^2-12z+45)`. Now lets complete the square under square root

`sqrt(-((z-6)^2-36+45))=sqrt(-((z-6)^2+9))`

Now lets do the variable change `z-6=3tanr`. With this change we have

`sqrt(-((z-6)^2+9))=sqrt(-(9tan^2r+9))=sqrt(-9)sqrt(tan^2r+1)`

and `dz=sec^2rdr`

We know that `sqrt(-9)=3i` and `tan^2r+1=sec^2r`

Now we have finally

`int1/(3isecr)sec^2rdr=int1/(3i)secrdr=1/(3i)intsecrdr`

Answer 2

`int \ 1/sqrt(-e^(2x)+12e^x-45) \ dx = i/(3sqrt(5)) ln|(6(sqrt((e^x-6)^2/9+1)-1))/(e^x-6)-sqrt(5)-1| - i/(3sqrt(5))ln|(6(sqrt((e^x-6)^2/9+1)-1))/(e^x-6)+sqrt(5)-1| + C`

Explanation:

We seek:

`I = int \ 1/sqrt(-e^(2x)+12e^x-45) \ dx`

The denominator is quadratic in `e^x`, so can complete the square to get:

`I = int \ 1/sqrt((-1)(e^(2x)-12e^x+45)) \ dx`
`\ \ = int \ 1/sqrt((-1)((e^(x)-6)^2-6^2+45)) \ dx`
`\ \ = int \ 1/(sqrt(-1)sqrt((e^(x)-6)^2-36+45)) \ dx`
`\ \ = int \ 1/(sqrt((e^(x)-6)^2+9)) * 1/i dx`
`\ \ = int \ 1/sqrt((e^(x)-6)^2+9) * 1/i * i/i dx`
`\ \ = -i \ int \ 1/sqrt((e^(x)-6)^2+9) dx`
`\ \ = -i \ I_1`, where `I_1 = int \ 1/sqrt((e^(x)-6)^2+9) dx`

Note that the denominator is positive `AA x in RR`, so we do not get a real solution.

To proceed with the integral, `I_1`, we can perform a substitution, Let:

`e^x-6 = 3tan theta => e^x = 3tan theta+6`

And if we differentiate wrt `x` we get:

`e^x = 3sec^2 theta (d theta)/dx => (d theta)/dx = (3tan theta+6)/(3sec^2 theta)`

And substituting into the integral `I_1` we get:

`I_1 = int \ 1/sqrt((3tan theta )^2+9) (3sec^2 theta)/(3tan theta + 6 ) \ d theta`

`\ \ \ = int \ 1/sqrt(9(tan^2 theta + 1)) (3sec^2 theta)/(3(tan theta + 2) ) \ d theta`

`\ \ \ = int \ 1/sqrt(9(sec^2theta)) (sec^2 theta)/((tan theta + 2) ) \ d theta`

`\ \ \ = 1/3 \ int \ (sec theta)/(tan theta + 2 ) \ d theta`

Next, we can perform a tangent half angle (Weierstrass substitution), so that:

`t=tan(theta/2) => theta=2arctant`

And we find that:

`sin theta=(2t)/(1+t^2)`, `cos theta = (1-t^2)/(1+t^2)` and `(d theta)/dt=2/(1+t^2)`

Thus we can perform a second substitution:

`I_1 = 1/3 \ int \ (1/cos theta)/((sin theta/cos theta) + 2 ) \ d theta`

`\ \ \ = 1/3 \ int \ ((1+t^2)/(1-t^2))/(((2t)/(1+t^2))((1+t^2)/(1-t^2)) + 2 ) \ 2/(1+t^2) \ dt`

`\ \ \ = 1/3 \ int \ (2/(1-t^2))/ ( (2t)/(1-t^2) + 2 ) \ dt`

`\ \ \ = 1/3 \ int \ (1/(1-t^2))/ ( (t + (1-t^2))/(1-t^2) ) \ dt`

`\ \ \ = -1/3 \ int \ 1/(t^2-t - 1) \ dt`

This integrand, is now significantly simplified, and we can use partial fraction decomposition (skipped) toi write in the form:

`I_1 = -1/3 2/sqrt(5){int 1/(2t-sqrt(5)-1) - 1/(2t+sqrt(5)-1) \ dt }`

Which we can now evaluate, to get:

`I_1 = -1/(3sqrt(5)){ln|2t-sqrt(5)-1| - ln|2t+sqrt(5)-1|}`

We can then restore the `t` substitution, using `t=tan(theta/2)`, to get:

`I_1 = -1/(3sqrt(5)){ln|2tan(theta/2)-sqrt(5)-1| - ln|2tan(theta/2)+sqrt(5)-1|}`

And we can restore the `theta` substitution using the relationship, , `theta = arctan(1/3(e^x-6))` from which we get:

`tan(theta/2) = (3(sqrt((e^x-6)^2/9+1)-1))/(e^x-6)`

And so we get:

`I_1 = -1/(3sqrt(5)){ln|(6(sqrt((e^x-6)^2/9+1)-1))/(e^x-6)-sqrt(5)-1| - ln|(6(sqrt((e^x-6)^2/9+1)-1))/(e^x-6)+sqrt(5)-1|}`

Thus:

`I = i/(3sqrt(5)) ln|(6(sqrt((e^x-6)^2/9+1)-1))/(e^x-6)-sqrt(5)-1| - i/(3sqrt(5))ln|(6(sqrt((e^x-6)^2/9+1)-1))/(e^x-6)+sqrt(5)-1| + C`