Question

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

Answer 1

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

Explanation:

`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

Please see below.

Explanation:

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`