How do you integrate #int 1/(sqrt(x^2-4))dx# using trigonometric substitution?

Question

4600 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-08T00:59:22

#int1/sqrt(x^2-4)dx = ln|x+sqrt(x^2-4)|+C#

#int1/sqrt(x^2-4)dx = int1/sqrt(4(x^2/4-1))dx#

#=1/2int1/sqrt((x/2)^2-1)dx#

Let #x/2 = sec(theta)#
Then #dx = 2sec(theta)tan(theta)#

#=>int1/sqrt(x^2-4)dx = 1/2int1/sqrt(sec^2(theta)-1)*2sec(theta)tan(theta)d theta#

#=int (sec(theta)tan(theta))/sqrt(tan^2(theta))d theta#

#=int sec(theta)d theta#

#=ln|sec(theta) + tan(theta)| + C#

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

(For the final equality, try drawing a right triangle with angle #theta# such that #sec(theta) = x/2# and solve for #tan(theta)#)

We could stop here, but we can also make this look a little nicer by getting rid of the #1/2#

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

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

But as #C# is an arbitrary constant, we can include the #ln(1/2)# in that to give us the final result

#ln|x+sqrt(x^2-4)|+C#