definite integral of 1 / sqrt[x^2 - 4] from 2 to 4

[degreeplus]

I need help finding the definite integral of 1/(x^2 - 4)^(1/2) from 2 to 4. I don't know if the antiderivative of the funcion is arcsin. Could some1 point me in the right direction?

 

[arthur ohlsten]

The definite integral is
ln[x+[x^2-4]^1/2 ] evaluated between 2 and 4

Arthur

 

[mark]

try setting x = 2sin(theta)
dx = 2cos(theta) dtheta

 

[arthur ohlsten]

Is the integral of dx/ [x^2-4]^1/2= ln[x+[x^2-4]^1/2 ?
let us take the derivative of ln[x+[x^2-4]^1/2]. the derivative is

1+1/2[x^2-4]^-1/2 [2x]
-----------------------------
x+[x2-4]^1/2

simplifying
[x^2-4]^1/2 + x
--------------------------
x[x^2-4]^1/2 +x^2-4

dividing numerator and denominator by x+[x^2-4]^1/2

1/[x^2-4]^1/2 proof

Arthur

 

[soroban]

Hello, degreeplus!

\(\displaystyle \L\int^{\;\;\;4}_2\frac{dx}{\sqrt{x^2\,-\,4}}\)

Let \(\displaystyle x\,=\,2\sec\theta\;\;\Rightarrow\;\;dx\,=\,2\sec\theta\tan\theta\,d\theta\)

. . and: \(\displaystyle \,\sqrt{x^2\,-\,4}\:=\:\sqrt{4\sec^2\theta\,-\,4}\:=\:\sqrt{4(\sec^2\theta\,-\,1)} \:=\:\sqrt{4\tan^2\theta}\:=\:2\tan\theta\)


Substitute: \(\displaystyle \L\:\int\frac{2\sec\theta\tan\theta\,d\theta}{2\tan\theta} \;=\;\int\sec\theta\,d\theta \;=\;\ln|\sec\theta\,+\,\tan\theta|\)


Back-substitute: \(\displaystyle \,\sec\theta\,=\,\frac{x}{2}\;\;\Rightarrow\;\;\tan\theta\,=\,\frac{\sqrt{x^2\,-\,4}}{2}\)


We have: \(\displaystyle \L\:\ln\left|\frac{x}{2}\,+\,\frac{\sqrt{x^2\,-\,4}}{2}\right|^4_2\)


Evaluate: \(\displaystyle \L\:\ln\left|\frac{4}{2}\,+\,\frac{\sqrt{4^2-4}}{2}\right|\:-\:\ln\left|\frac{2}{2}\,+\,\frac{\sqrt{2^2-4}}{2}\right|\)

. . \(\displaystyle \L=\;\ln\left(2\,+\,\frac{\sqrt{12}}{2}\right)\:-\:\ln(1 + 0)\;=\;\fbox{\ln\left(2\,+\,\sqrt{3}\right)}\)

 

[skeeter]

I agree with Soroban's excellent solution, but technically speaking, this is an improper integral, is it not?

\(\displaystyle \L \lim_{a\rightarrow 2^+} \int_a^4 \frac{1}{\sqrt{x^2-4}} dx\)