How do you integrate #int 2/sqrt(x^2-4)dx# using trigonometric substitution?

1 Answer
Jun 28, 2016

Use x = 2*sec(u) as a first substitution.

Explanation:

With #x = 2*sec(u), therefore dx = 2*sec(u)*tan(u)#

#sqrt(x^2 - 4) = sqrt(4*sec^2(u) - 4) = 2*sqrt(sec^2(u) - 1)#

From #sin^2(a) + cos^2(a) = 1#, divide through by #cos^2(a)# to obtain

#tan^2(a) + 1 = sec^2(a)#

So, we have:

#int 2/(sqrt(x^2-4) dx# # = int (4*sec(u)*tan(u))/(2sqrt(tan^2(u))# # du #

So, integral of # 2*int sec(u) du#

Multiply numerator and denominator by #sec(u) + tan(u)# to obtain

# 2*int (sec^2(u) + sec(u)*tan(u))/(sec(u) + tan(u)) du#

Use substitution #r = sec(u) + tan(u)# which gives

# 2*int (dr)/r#

# = 2*ln(r) + C#

# = 2*ln(sec(u) + tan(u)) + C#

# = 2*ln(sec(arcsec(x/2) + tan(arcsec(x/2)) + C#

# = 2*ln(1/2*(x + sqrt(x^2 - 4))) + C#

And there we have it, let me know if you need any help with the last bit but it's fairly easy to figure out if you look at a few triangles.