How do you integrate #int 1/(x^2+25)# by trigonometric substitution?

1 Answer
mason m
Sep 6, 2016

#1/5arctan(x/5)+C#

Explanation:

We have:

#I=intdx/(x^2+25)#

Since we know that #tan^2x+1=sec^2x#, we will make the denominator similar to that. Let #x=5tantheta#. Note that this also implies that #dx=5sec^2thetad theta#. Thus:

#I=int(5sec^2theta)/(25tan^2theta+25)d theta#

#I=1/5int(sec^2theta)/(tan^2theta+1)d theta#

#I=1/5intsec^2theta/sec^2thetad theta#

#I=1/5intd theta#

#I=1/5theta#

From #x=5tantheta#, we see that #theta=arctan(x/5)#:

#I=1/5arctan(x/5)+C#

Note that this also acts in accordance with the rule:

#int1/(x^2+a^2)=1/a arctan(x/a)+C#

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