How do you integrate #int (4x)/sqrt(x^2-49)dx# using trigonometric substitution?

Question

Charles Anctil · Accepted Answer

#I = 4(x^2 - 49)^(1/2) +C#

Trig substitution can be used, but is not needed to solve this problem. If we let #u = x^2 - 49#, we see that #du = 2xdx# and #dx = (du)/(2x)#.

#I = int (4x)/sqrt(u) * du/(2x)#

#I = 2int 1/sqrt(u)du#

#I = 2int u^(-1/2) du#

#I = 2(2u^(1/2)) + C#

#I = 4(x^2 - 49)^(1/2) +C#

Hopefully this helps!

Scarlett Allison · Answer

undefined To integrate $ \int \frac{4x}{\sqrt{x^2 - 49}} \, dx $ using trigonometric substitution, we can let $ x = 7\sec(\theta) $. Then, $ dx = 7\sec(\theta)\tan(\theta) \, d\theta $.

Substituting these expressions into the integral, we get:

$$ \int \frac{4 \cdot 7\sec(\theta)\tan(\theta)}{\sqrt{(7\sec(\theta))^2 - 49}} \, d\theta $$

$$ = \int \frac{28\sec(\theta)\tan(\theta)}{\sqrt{49\sec^2(\theta) - 49}} \, d\theta $$

$$ = \int \frac{28\sec(\theta)\tan(\theta)}{\sqrt{49(\sec^2(\theta) - 1)}} \, d\theta $$

$$ = \int \frac{28\sec(\theta)\tan(\theta)}{\sqrt{49\tan^2(\theta)}} \, d\theta $$

$$ = \int \frac{28\sec(\theta)\tan(\theta)}{7\tan(\theta)} \, d\theta $$

$$ = 4\int d\theta $$

$$ = 4\theta + C $$

Finally, substituting back $ \theta = \sec^{-1}(\frac{x}{7}) $, we have:

$$ = 4\sec^{-1}\left(\frac{x}{7}\right) + C $$

So, $ \int \frac{4x}{\sqrt{x^2 - 49}} \, dx = 4\sec^{-1}\left(\frac{x}{7}\right) + C $.