How do you integrate #int dx/sqrt(x^2-64)# using trig substitutions?

Question

Scarlett Allison · Accepted Answer

# I = ln | sqrt(x^2/64 -1 ) + x/8 | + C #

We seek: # I = int \ 1/sqrt(x^2-64) \ dx # We can perform the substitution: # x = 8sec theta => sec theta = x/8 # And differentiating wrt #theta#: # dx/(d theta) = 8sec theta tan theta # And Substituting into the integral, it becomes: # I = int \ 1/sqrt((8sec theta)^2-64) \ 8sec theta tan theta \ d theta# # \ \ = int \ (8sec theta tan theta)/sqrt(64sec^2 theta-64) \ d theta# # \ \ = int \ (sec theta tan theta)/sqrt(sec^2 theta-1) \ d theta# # \ \ = int \ (sec theta tan theta)/sqrt(tan^2 theta) \ d theta# # \ \ = int \ sec theta \ d theta# # \ \ = ln | tan theta + sec theta | + C # And using the identity: # 1 + tan^2 A -= sec^2 A => tan theta = sqrt(sec^2 theta -1 ) # Allowing us to restore the earlier substitution: # I = ln | sqrt(sec^2 theta -1 ) + sec theta | + C # # \ \ = ln | sqrt((x/8)^2 -1 ) + x/8 | + C # # \ \ = ln | sqrt(x^2/64 -1 ) + x/8 | + C #

Christopher Allers · Answer

undefined To integrate $\int \frac{{dx}}{{\sqrt{x^2 - 64}}}$ using trigonometric substitution, you can let $x = 8\sec(\theta)$, where $\sec(\theta) = \frac{1}{\cos(\theta)}$. Then, $dx = 8\sec(\theta)\tan(\theta)d\theta$.

Substituting $x = 8\sec(\theta)$ and $dx = 8\sec(\theta)\tan(\theta)d\theta$ into the integral, you get:

$$\int \frac{{8\sec(\theta)\tan(\theta)}}{{\sqrt{(8\sec(\theta))^2 - 64}}}d\theta$$

Simplify the expression under the square root:

$$\sqrt{(8\sec(\theta))^2 - 64} = \sqrt{64\tan^2(\theta)} = 8\tan(\theta)$$

So, the integral becomes:

$$\int \frac{{8\sec(\theta)\tan(\theta)}}{{8\tan(\theta)}}d\theta$$

Simplify further:

$$\int d\theta$$

Integrating $d\theta$ with respect to $\theta$ gives:

$$\theta + C$$

Finally, substitute back $\theta = \sec^{-1}(\frac{x}{8})$ into the result to obtain the final answer.