How do you integrate #intdx/ sqrt(x^2 - a^2)#?

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Answer 1 · 2018-04-19T22:08:45

#intdx/sqrt(x^2-a^2)="arcosh"(x/a)+"c"#

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

Now, let #u=x/a# and #du=1/adx#

Then

#int1/(asqrt((x/a)^2-1))dx=int1/sqrt(u^2-1)dx="arcosh"(u)+"c"="arcosh"(x/a)+"c"#

Answer 2 · 2018-04-19T23:49:05

#int dx/sqrt(x^2-a^2) = ln abs (x +sqrt(x^2-a^2)) +C#

The integrand is defined for #x in (-oo,-a) uu (a,+oo)#. Let us focus first on #x in (a,+oo)# and substitute:

#x = a sect#

#dx = asect tant#

with #t in (0,pi/2)#

so:

#int dx/sqrt(x^2-a^2) = int (asect tant dt)/sqrt(a^2sec^2t-a^2)#

#int dx/sqrt(x^2-a^2) = int (sect tant dt)/sqrt(sec^2t-1)#

Use now the trigonometric identity:

#sec^2t-1 = tan^2t#

and as for #t in (0,pi/2)# the tangent is positive:

#sqrt(sec^2t-1) = tan^#

Then:

#int dx/sqrt(x^2-a^2) = int (sect tant dt)/tant#

#int dx/sqrt(x^2-a^2) = int sect dt#

#int dx/sqrt(x^2-a^2) = ln abs (sect +tant) +C#

Undoing the substitution:

#int dx/sqrt(x^2-a^2) = ln abs (x/a +sqrt(x^2/a^2-1)) +C#

#int dx/sqrt(x^2-a^2) = ln abs (x +sqrt(x^2-a^2)) +C#

By differentiating we can see that the solution is valid also for #x in (-oo,-3)#