How do i integrate: square root of((x^2)-4) ?

[sara_87]

how do i integrate: square root of((x^2)-4) i know i have to use substitution but what do i substitute?

 

[HallsofIvy]

Well, I remember that $sin^2(\theta)+ cos^2(\theta)= 1$ and if I divide both sides of that by [/itex]cos^2(\theta)[/itex], I get $tan^2(\theta)+ 1= sec^2(\theta)$ and from that, $tan^2(\theta)= sec^2(\theta)- 1$. That tells me that if I factor out a "4" from that squareroot, getting $2\sqrt{(x/2)^2- 1}$, I can the part inside the square root into a perfect square (and so get rid of the square root) by making the substitution $x/2= sec(\theta)$.

 

[sara_87]

i used the substitution x/2=secu then i reached a point where i got stuck again:
4*integral(1/(cosu)^3) du -4ln(secu+tanu)
so how do i integrate (1/(cosu)^3)??
thank u.

 

[bumfluff]

Maybe (sec u)^2(sec u) would help and then maybe integrate by parts, but that probably won't work in the end either.

 

[Big-T]

You could try expanding the fraction by cos u, then using cos^2 u = 1-sin^2, substitution and partial fractions.

 

[arildno]

Alternatively, use the substitution x=2Cosh(u)

 

[HallsofIvy]

That is cosine to an odd power. "Factor" out a cosine to use with the du and convert the rest (even power) to sine.

Here, the cosine is in the denominator so "factoring" out a cosine gives
$$\int \frac{cos(u)du}{cos^4(u)}= \int \frac{cos(u)du}{(1- sin^2(u))^2}$$
Now let v= sin(u) so dv= cos(u)du and you have
$$\int \frac{dv}{(1-v^2)^2}= \int \frac{dv}{(1-v)^2(1+v)^2}$$
which can be done by partial fractions.