How do you use substitution to integrate # 1/sqrt(x^2 -4)#?

Question

Charles Anctil · Accepted Answer

ln I#x/2 +sqrt(x^2-4)/2#I +C

Let x= 2sec u. so that dx= 2sec u tan u du. The given integral becomes #int 1/sqrt (4sec^2 u -4) 2 sec u tan u du# = #int secu du# = ln lsecu +tan u I +C = ln I#x/2 + sqrt(x^2-4)/2#I +C

Charles Anctil · Answer

undefined To integrate $ \frac{1}{\sqrt{x^2 -4}} $ using substitution, we can use the trigonometric substitution method. Let $ x = 2\sec(\theta) $, then $ dx = 2\sec(\theta)\tan(\theta) \, d\theta $.

Substituting $ x = 2\sec(\theta) $ into $ \sqrt{x^2 -4} $, we get $ \sqrt{(2\sec(\theta))^2 - 4} = \sqrt{4\tan^2(\theta)} = 2\tan(\theta) $.

Thus, $ \frac{1}{\sqrt{x^2 -4}} = \frac{1}{2\tan(\theta)} $.

Now, substitute $ dx = 2\sec(\theta)\tan(\theta) \, d\theta $ and $ \frac{1}{\sqrt{x^2 -4}} = \frac{1}{2\tan(\theta)} $ into the integral:

$ \int \frac{1}{\sqrt{x^2 -4}} \, dx = \int \frac{1}{2\tan(\theta)} \cdot 2\sec(\theta)\tan(\theta) \, d\theta $.

This simplifies to:

$ \int d\theta $.

Now, integrating $ \int d\theta $ yields $ \theta + C $.

Finally, substitute back $ x = 2\sec(\theta) $ into $ \theta + C $ to get the final answer in terms of $ x $.

HIX Tutor · Answer

undefined Sure! Here’s how to use substitution to integrate $ \frac{1}{\sqrt{x^2 - 4}} $:

1. **Identify the substitution**: We want to use $ x = 2 \sec(\theta) $. This is because $ \sec^2(\theta) - 1 = \tan^2(\theta) $.

2. **Find $ dx $**: Differentiate $ x = 2 \sec(\theta) $:
   $$
   dx = 2 \sec(\theta) \tan(\theta) d\theta
   $$

3. **Rewrite the integral**: Substitute $ x $ and $ dx $ in the integral.
   $$
   \sqrt{x^2 - 4} = \sqrt{(2 \sec(\theta))^2 - 4} = \sqrt{4 \sec^2(\theta) - 4} = \sqrt{4(\sec^2(\theta) - 1)} = \sqrt{4 \tan^2(\theta)} = 2 \tan(\theta)
   $$
   So, the integral becomes:
   $$
   \int \frac{1}{2 \tan(\theta)} \cdot 2 \sec(\theta) \tan(\theta) d\theta
   $$
   This simplifies to:
   $$
   \int \sec(\theta) d\theta
   $$

4. **Integrate**: The integral of $ \sec(\theta) $ is:
   $$
   \ln | \sec(\theta) + \tan(\theta) | + C
   $$

5. **Back-substitute**: Return to $ x $ using $ \theta $:
   $$
   \sec(\theta) = \frac{x}{2}, \quad \tan(\theta) = \frac{\sqrt{x^2 - 4}}{2}
   $$
   Thus, the final answer is:
   $$
   \ln \left| \frac{x}{2} + \frac{\sqrt{x^2 - 4}}{2} \right| + C
   $$

And that's how you do it!