Question

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

Answer 1

`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`