How do you integrate #int 1/sqrt(9x^2-18x+18) # using trigonometric substitution?

Question

William Abdalla · Accepted Answer

#=1/3sinh^-1(x-1)+C#

First of all, we complete the square/ re write in vertex form the quadratic under the square root:

#9x^2-18x+18# #=9(x^2-2)+18# #=9(x-1)^2+9#

So the integral can be re written as:

#int1/sqrt(9(x-1)^2+9)dx=1/3int1/sqrt((x-1)^2+1)dx#

(Note the 9 under the square root has been factored out to the front)

Now consider the substitution #sinh(u)=x-1# It will follow that #cosh(u)du=dx#

Now substitute this into the integral to get:

#1/3intcosh(u)/sqrt(sinh^2(u)+1)dx#

Use the identity: #cosh^2(u)-sinh^2(u)=1# to re write the bottom as:

#1/3intcosh(u)/sqrt(cosh^2(u))dx=1/3intcosh(u)/cosh(u)du = 1/3intdu#

Now evaluating the integral and reversing the substitution:

#1/3u +C# #=1/3sinh^-1(x-1)+C#

William Abdalla · Answer

undefined To integrate $ \int \frac{1}{\sqrt{9x^2 - 18x + 18}} $ using trigonometric substitution, follow these steps:

1. Complete the square inside the square root to make it easier to work with.
   $$ 9x^2 - 18x + 18 = 9(x^2 - 2x + 2) = 9[(x - 1)^2 + 1] $$

2. Now, substitute $ x - 1 = \sqrt{3}\tan(\theta) $, which implies $ x = 1 + \sqrt{3}\tan(\theta) $, and $ dx = \sqrt{3}\sec^2(\theta)d\theta $.

3. Substitute $ x $ and $ dx $ in terms of $ \theta $ in the integral.
   $$ \int \frac{1}{\sqrt{9[(x - 1)^2 + 1]}} dx = \int \frac{1}{\sqrt{9(1 + \tan^2(\theta))}} \cdot \sqrt{3}\sec^2(\theta) d\theta $$

4. Simplify inside the square root and outside the integral.
   $$ \int \frac{\sqrt{3}\sec^2(\theta)}{\sqrt{9\sec^2(\theta)}} d\theta = \int \frac{\sqrt{3}\sec(\theta)}{3} d\theta $$

5. Now integrate $ \sqrt{3}\sec(\theta)/3 $ with respect to $ \theta $.
   $$ \int \frac{\sqrt{3}\sec(\theta)}{3} d\theta = \frac{\sqrt{3}}{3} \int \sec(\theta) d\theta $$

6. Finally, integrate $ \sec(\theta) $ with respect to $ \theta $.
   $$ \frac{\sqrt{3}}{3} \ln|\sec(\theta) + \tan(\theta)| + C $$

7. Substitute back $ \theta $ in terms of $ x $.
   $$ \frac{\sqrt{3}}{3} \ln|\sec(\arctan(\frac{x-1}{\sqrt{3}})) + \tan(\arctan(\frac{x-1}{\sqrt{3}}))| + C $$

8. Simplify further if necessary.