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

Question

Ezra Adams · Accepted Answer

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#

Evelyn Agard · Answer

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#

Charles Arocha · Answer

undefined To integrate $ \int \frac{dx}{x^2 + 25} $ using trigonometric substitution, we let $ x = 5 \tan(\theta) $, which implies $ dx = 5 \sec^2(\theta) d\theta $. Substituting these expressions into the integral, we get:

$$ \int \frac{5 \sec^2(\theta) d\theta}{25\tan^2(\theta) + 25} $$

Simplify the expression:

$$ \int \frac{5 \sec^2(\theta) d\theta}{25(\tan^2(\theta) + 1)} $$
$$ = \int \frac{5 \sec^2(\theta) d\theta}{25\sec^2(\theta)} $$
$$ = \frac{1}{5} \int d\theta $$

Integrating $ d\theta $ gives:

$$ \frac{1}{5} \theta + C $$

Finally, we substitute back for $ \theta $ using the original substitution $ x = 5 \tan(\theta) $. Since $ \tan(\theta) = \frac{x}{5} $, we can find $ \theta $ as $ \arctan(\frac{x}{5}) $.

Therefore, the final result is:

$$ \frac{1}{5} \arctan\left(\frac{x}{5}\right) + C $$