How do you integrate #int 1/sqrt(4x^2-12x+34) # using trigonometric substitution?

Question

Christopher Allers · Accepted Answer

Integration without trigonometric substitution.
#I=1/2ln|(2x-3)+sqrt(4x^2-12x+34)|+c# Here, #I=int1/sqrt(4x^2-12x+34)dx# #=int1/sqrt(4x^2-12x+9+25)dx# #=int1/sqrt((2x-3)^2+5^2)dx# Subst, #color(blue)(2x-3=u)=>2dx=du=>dx=1/2du# So, #I=int1/sqrt(u^2+5^2)xx1/2du# #=1/2color(red)(int1/sqrt(u^2+5^2)du# #=1/2color(red)(ln|u+sqrt(u^2+5^2)|+c# Subst. back ,#color(blue)(u=2x-3 # ,we get #I=1/2ln|(2x-3)+sqrt((2x-3)^2+5^2)|+c# #=1/2ln|(2x-3)+sqrt(4x^2-12x+34)|+c# ......................................................................................... Note : #color(red)((1)int1/sqrt(x^2+a^2)dx=ln|x+sqrt(x^2+a^2)|+c#

William Abdalla · Answer

#=1/2ln|sqrt(4x^2-12x+34)+(2x-3)|+C #

Here,

#I=int1/sqrt(4x^2-12x+34)dx#

#=int1/sqrt(4x^2-12x+9+25)dx#

#=int1/sqrt((2x-3)^2+5^2)dx#

Subst, #2x-3=5tanu=>2dx=5sec^2udu#

#=>dx=5/2sec^2udu and color(blue)(tanu=(2x-3)/5#

So

#I=int1/sqrt(5^2tan^2u+5^2)xx5/2sec^2udu#

#=5/2intsec^2u/(5secu)du#

#=1/2intsecudu#

#=1/2ln|secu+tanu|+c#

#=1/2ln|sqrt(tan^2u+1)+tanu|+c#

Subst. back , #color(blue)(tanu=(2x-3)/5# we get

#I=1/2ln|sqrt(((2x-3)/5)^2+1)+((2x-3)/5)|+c#

#=1/2ln|(sqrt(4x^2-12x+34))/5+(2x-3)/5|+c#

#=1/2ln|((sqrt(4x^2-12x+34))+(2x-3))/5|+c#

#=1/2ln|sqrt(4x^2-12x+34)+(2x-3)|-1/2ln5+c#

#=1/2ln|sqrt(4x^2-12x+34)+(2x-3)|+C #

#where,C=-1/2ln5+c#

Emilia Aguilar · Answer

undefined Let $ x = \frac{3}{2} + \frac{1}{2}\sqrt{3}\sec(\theta) $. Then $ dx = \frac{1}{2}\sqrt{3}\sec(\theta)\tan(\theta)\, d\theta $.
Substitute for x and dx in the integral.
$ \int \frac{1}{\sqrt{4x^2 - 12x + 34}} \, dx = \int \frac{1}{\sqrt{3}\tan(\theta)\sqrt{3}\sec(\theta)} \cdot \frac{1}{2}\sqrt{3}\sec(\theta)\tan(\theta)\, d\theta $.
Simplify the integrand.
$ = \int \frac{1}{2\sin(\theta)} \, d\theta $.
Use the identity $ \csc(\theta) = \frac{1}{\sin(\theta)} $ to simplify the integral.
$ = \int \frac{1}{2}\csc(\theta) \, d\theta $.
Integrate $ \csc(\theta) $ with respect to $ \theta $.
$ = -\frac{1}{2}\ln|\csc(\theta) + \cot(\theta)| + C $.
Substitute back for $ \theta $ in terms of $ x $.
$ = -\frac{1}{2}\ln\left|\frac{1}{\sin(\theta)} + \frac{\cos(\theta)}{\sin(\theta)}\right| + C $.
Simplify the expression inside the absolute value.
$ = -\frac{1}{2}\ln\left|\frac{1+\cos(\theta)}{\sin(\theta)}\right| + C $.
Use the Pythagorean identity $ 1 + \cos(\theta) = \csc(\theta) $ to simplify further.
$ = -\frac{1}{2}\ln\left|\csc(\theta)\csc(\theta)\right| + C $.
Simplify the expression inside the absolute value.
$ = -\frac{1}{2}\ln\left|\csc^2(\theta)\right| + C $.
Use the property of logarithms $ \ln(a^b) = b\ln(a) $ to simplify further.
$ = -\ln|\csc(\theta)| + C $.
Finally, substitute $ \csc(\theta) = \sqrt{4x^2 - 12x + 34} $ back in.
$ = -\ln\left|\sqrt{4x^2 - 12x + 34}\right| + C $.
Simplify the expression.
$ = -\frac{1}{2}\ln|4x^2 - 12x + 34| + C $.