How do you integrate #int 1/sqrt(x^2-6) # using trigonometric substitution?

Question

Scarlett Allison · Accepted Answer

#int\frac{1}{\sqrt{x^2-6}}=log_e((x+sqrt{x^2-6})/\sqrt6)+c#

Given equation is #int\frac{1}{\sqrt{x^2-6}}#

For this equation, we'll substitute #x=\sqrt{6}sec(t)# So, #dx=\sqrt{6}sec(t)tan(t)dt#

Substituting these values into the equation, we get #int\frac{\sqrt{6}sec(t)tan(t)dt}{\sqrt{6sec^2(x)-6}}# Taking #\sqrt{6}# common from the denominator, we end with #int\frac{sec(t)tan(t)dt}{\sqrt{sec^2(t)-1}}#

I believe you're well friends with this general trigonometric equation #tan^2(\theta)+1=sec^2(\theta)# So subtracting #1# from both sides, we get #tan^2(\theta)=sec^2(\theta)-1#

Therefore, the denominator of the integral becomes #int\frac{sec(t)tan(t)dt}{tan(t)}# Cancelling #tan(t)# function, we end with #intsec(t)dt=log_e(sec(t)+tan(t))+c#

Now, we know that #x=\sqrt{6}sec(t)\impliessec(t)=x/\sqrt{6}# Also, from the above trigonometric identity, #tan^2(t)=x^2/6-1=(x^2-6)/6\impliestan(t)=\sqrt{\frac{x^2-6}{6}}#

So substituting the above values for their respective functions, we get the integral of the said main function.

Emma Aldridge · Answer

undefined To integrate $\int \frac{1}{\sqrt{x^2 - 6}}$ using trigonometric substitution, we follow these steps:

1. Recognize that the expression under the square root resembles a trigonometric identity, particularly $\sec^2(\theta) - 1 = \tan^2(\theta)$.
2. Substitute $x = \sqrt{6}\sec(\theta)$, which implies $dx = \sqrt{6}\sec(\theta)\tan(\theta)d\theta$.
3. Rewrite the integral in terms of $\theta$ using the substitution.
4. Solve the integral in terms of $\theta$.
5. Substitute back $x$ in terms of $\theta$.
6. Simplify the result.

After substitution, the integral becomes:

$$
\int \frac{1}{\sqrt{6\sec^2(\theta) - 6}} \cdot \sqrt{6}\sec(\theta)\tan(\theta)d\theta
$$

Simplify this expression to:

$$
\int \frac{1}{\sqrt{6\tan^2(\theta)}} \cdot \sqrt{6}\sec(\theta)\tan(\theta)d\theta
$$

This simplifies further to:

$$
\int \frac{1}{\sqrt{6}\tan(\theta)} \cdot \sqrt{6}\sec(\theta)\tan(\theta)d\theta
$$

The $\sqrt{6}\tan(\theta)$ terms cancel out, leaving:

$$
\int d\theta
$$

This integral evaluates to $\theta + C$. Finally, substitute back $x = \sqrt{6}\sec(\theta)$, so $\theta = \arccos\left(\frac{x}{\sqrt{6}}\right)$.

Thus, the result is $\arccos\left(\frac{x}{\sqrt{6}}\right) + C$.