Question

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

Answer 1

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

Explanation:

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.