How do you find the integral of #5/[sqrt(9x^2-16)] dx#?

Question

Scarlett Allison · Accepted Answer

The answer is #=5/3ln(|3/4x+sqrt((3/4x)^2-1)|)+C#

Perform some simplification

#5/sqrt(9x^2-16)=5/(4sqrt (((3/4x)^2)-1))#

Let #3/4x=secu#, #=>#, #3/4dx=secutanudu#

#sqrt((3/4x)^2-1)=sqrt(sec^2u-1)=tanu#

Therefore, the integral is

#I=int(5dx)/(sqrt(9x^2-16))=5/4int(4/3secutanudu)/(tanu)#

#=5/3intsecudu#

#=5/3int(secu(secu+tanu)du)/(secu+tanu)#

Let #v=secu+tanu#, #=>#, #dv=(sec^2u+secutanu)du#

Therefore,

#I=5/3int(dv)/(v)#

#=5/3ln(v)#

#=5/3ln(secu+tanu)#

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

Samantha Adkison · Answer

undefined To find the integral of $ \frac{5}{\sqrt{9x^2-16}} $ with respect to $ x $, you can use the trigonometric substitution method. Let $ x = \frac{4}{3} \sec(\theta) $. Then $ dx = \frac{4}{3} \sec(\theta) \tan(\theta) d\theta $. Substituting these into the integral, it simplifies to $ \int \frac{20\sec(\theta)\tan(\theta)}{\sqrt{16\sec^2(\theta)}} d\theta $. This simplifies further to $ \int \frac{20\tan(\theta)}{4|\sec(\theta)|} d\theta $. Since $ |\sec(\theta)| = \sec(\theta) $ for $ \theta $ in the domain of the integral, this becomes $ \int \frac{5\tan(\theta)}{\sec(\theta)} d\theta $. Using the identity $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $ and $ \sec(\theta) = \frac{1}{\cos(\theta)} $, the integral becomes $ \int 5\sin(\theta) d\theta $. Integrating $ \sin(\theta) $ yields $ -5\cos(\theta) + C $, where $ C $ is the constant of integration. Finally, substituting back $ \cos(\theta) = \frac{4}{3x} $ gives the result $ \int \frac{5}{\sqrt{9x^2-16}} dx = -\frac{5}{3}\sqrt{9x^2-16} + C $, where $ C $ is the constant of integration.