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

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#

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#