How do you integrate #int dx/(x^2+25)# using trig substitutions?

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Answer 1 · 2016-12-08T20:20:40

Let #x = 5tan(theta)#, then #dx = sec^2(theta)d"theta#. To reverse the substitution, use #theta = tan^-1(x/5)#

#intdx/(x^2 + 25)#

Let #x = 5tan(theta)#, then #dx = sec^2(theta)d"theta#:

#intsec^2(theta)/((5tan(theta))^2 + 25)d"theta = #

#intsec^2(theta)/(25tan^2(theta) + 25)d"theta = #

#1/25intsec^2(theta)/(tan^2(theta) + 1)d"theta =#

Use the identity #sec^2(theta) = tan^2(theta) + 1#

#1/25intsec^2(theta)/sec^2(theta)d"theta =#

#1/25intd"theta =#

#1/25theta+ C =#

Reverse the substitution; substitute #tan^-1(x/5) " for "theta#

#1/25tan^-1(x/5)+ C#

Answer 2 · 2016-12-08T20:30:16

Please see below.

A good way to sort out trig substitutions is to recall the pythagorean identities in trig.

#1-sin^2x = cos^2x# is useful if we see #a-bu^2#. (We try to get #k(1-sin^2theta) = k cos^2 theta#)

In this case, we have a square plus a number. We think about how to use #trig^2 + 1#

Eventually (perhaps by going through the list of pythagorean identities) we recall that

#tan^2theta + 1 = sec^2 theta#. That gives us our substitution.

We want to have #25tan^2 theta + 25#, so we'll use #x = 5tan^2 theta#. The denominator becomes

#(5tan theta)^2 + 25 = 25(tan^2 theta + 1) = 25(sec^2 theta)#

We also get #dx = 5sec^2 theta d theta#, so out integral becomes

#int (dx)/(x^2+25) = int (5 sec^2 theta d theta)/(25sec^2 theta)#

# = 1/5 int d theta = 1/5 theta + C#

Since #x = 5 tan theta#, we have #theta = tan^-1 (x/5) + C#

So

#int (dx)/(x^2+25) = 1/5 tan^-1(x/5) + C#