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

1 Answer
Jan 25, 2016

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.