How do you integrate #int 1/sqrt(x^2-9)# by trigonometric substitution?

3 Answers
Dec 11, 2016

#int dx/sqrt(x^2-9) = log |x/3+1/3sqrt(x^9-9)|#

Explanation:

Substitute:

#x=3sect#
#dx = 3sect tant dt#

and you have for #t in (0,pi/2)#:

#int dx/sqrt(x^2-9) = int (3sect tant dt)/sqrt(9sec^2t-9)= int (sect tant dt)/sqrt(1/cos^2t-1) = int (sect tant dt)/sqrt((1-cos^2t)/cos^2t) = int (sect tant dt)/sqrt(sin^2t/cos^2t) =int (sect tant dt)/sqrt(tan^2t)#

In this interval for #tant > 0#, so:

#int dx/sqrt(x^2-9) = int (sect tant dt)/tant =intsectdt= log |sect+ tant|#

To go back to x we note that:

#sect = x/3#

and #tant= sint*sect=sectsqrt(1-1/sec^t2t)= x/3sqrt(1-9/x^2)=1/3sqrt(x^9-9)#.

Finally:

#int dx/sqrt(x^2-9) = log |x/3+1/3sqrt(x^9-9)|#

Dec 11, 2016

Given: #int 1/sqrt(x^2 - 9)dx#

Let #x = 3sec(u)", then "dx = 3tan(u)sec(u)du#

Substitute this into the integral:

#int (3tan(u)sec(u))/sqrt((3sec(u))^2 - 9)du = #

#int (3tan(u)sec(u))/sqrt(9sec^2(u) - 9)du = #

#int (3tan(u)sec(u))/(3sqrt(sec^2(u) - 1))du = #

#int (tan(u)sec(u))/(sqrt(sec^2(u) - 1))du = #

#int (tan(u)sec(u))/(sqrt(tan^2(u)))du = #

#int (tan(u)sec(u))/tan(u)du = #

This integral is in any good list of integrals:

#int sec(u)du = ln(tan(u) + sec(u)) + C#

#int 1/sqrt(x^2 - 9)dx = ln(sqrt(sec^2(u) - 1) + sec(u)) + C#

Substitute #x/3# for #sec(u)#:

#int 1/sqrt(x^2 - 9)dx = ln(sqrt(x^2/9 - 1) + x/3) + C#

Dec 11, 2016

#cosh^{-1)|x/3|+C#

Explanation:

Substitute #x=3cosh u#, #dx=3sinh u du# and use #cosh^2u -sinh^2u = 1# to get the integral as being #u#, which is #cosh^{-1}(x/3)#
which can be written as #ln|x+sqrt(x^2-9)|+C# which is equivalent to
#ln |(x/3+(1/3)sqrt(x^2-)9)|# given above, as the two answers differ by a constant of #ln 3# which is subsumed into the constant of integration.