How do you integrate #int x/sqrt(x^2-25)# by trigonometric substitution?

Question

Christopher Allers · Accepted Answer

#int x/sqrt(x^2-25)dx = sqrt(x^2-25) + C#

You do not need a trigonometric substitution to solve this integral.

Substituting:

#t=x^2-25# #dt= 2xdx#

you have:

#int x/sqrt(x^2-25)dx = 1/2 int dt/sqrt(t) = sqrt(t)+C#

and substituting back #x#:

#int x/sqrt(x^2-25)dx = sqrt(x^2-25) + C#

Scarlett Allison · Answer

As has been shown, trigonometric substitution is a waste of time and effort, but it can be used with the substitution #x=5sectheta=>dx=5secthetatanthetad theta#. #intx/sqrt(x^2-25)dx=int(5sectheta)/sqrt(25sec^2theta-25)(5secthetatanthetad theta)# Note that #sqrt(25sec^2theta-25)=5sqrt(sec^2theta-1)=5tantheta# via the form of the Pythagorean identity #tan^2theta+1=sec^2theta#. #=int(5sectheta)/(5tantheta)(5secthetatanthetad theta)=5intsec^2thetad theta=5tantheta+C# Note that #tantheta=sqrt(sec^2theta-1)# and from our original substitution #sectheta=x/5#: #=5sqrt(sec^2theta-1)+C=5sqrt(x^2/25-1)+C=5sqrt((x^2-25)/25)+C# #=sqrt(x^2+25)+C#

Emily Ammerman · Answer

undefined To integrate $\frac{x}{\sqrt{x^2 - 25}}$ by trigonometric substitution, you can use the substitution $x = 5\sec(\theta)$, where $\theta$ is the angle in a right triangle with adjacent side $5$ and hypotenuse $x$.

With this substitution, you can express $dx$ as $5\sec(\theta)\tan(\theta)d\theta$ and $\sqrt{x^2 - 25}$ as $5\tan(\theta)$. Substituting these expressions into the integral, you get:

$$ \int \frac{x}{\sqrt{x^2 - 25}} dx = \int \frac{5\sec(\theta)\tan(\theta)}{5\tan(\theta)} \times 5\sec(\theta)\tan(\theta) d\theta $$

$$ = \int 5\sec^2(\theta) d\theta $$

Now, integrate $\sec^2(\theta)$ with respect to $\theta$ to get:

$$ = 5\tan(\theta) + C $$

Finally, revert back to the variable $x$ using the relation $x = 5\sec(\theta)$ and $\tan(\theta) = \frac{\sqrt{x^2 - 25}}{5}$:

$$ = 5\tan(\theta) + C = 5\left(\frac{\sqrt{x^2 - 25}}{5}\right) + C = \sqrt{x^2 - 25} + C $$

So, the integral of $\frac{x}{\sqrt{x^2 - 25}}$ by trigonometric substitution is $\sqrt{x^2 - 25} + C$, where $C$ is the constant of integration.