Question

How do you integrate `int dx/sqrt(9x^2+4)` using trig substitutions?

Answer 1

`int \ 1/sqrt(9x^2+4) \ dx = 1/3 \ ln (sqrt(1+(9x^2)/4) + (3x)/2)+c`

Explanation:

We want to find:

`I = int \ 1/sqrt(9x^2+4) \ dx`

For the integrand `1/sqrt(9x^2+4)` a suitable substitution should be:

`tanu = (3x)/2 => tan^2u = (9x^2)/4`
`" "=> 9x^2=4tan^2u`

And differentiating wrt `x` we get:

`sec^2u(du)/dx=3/2 => 2/3sec^2u(du)/dx=1`

Applying the substitution along with `1+tan^2A-=sec^2A` we get;

`I = int \ (2/3sec^2u)/sqrt(4tan^2u+4) \ du`
`\ \ = 2/3 \ int \ (sec^2u)/sqrt(4(tan^2u+1)) \ du`
`\ \ = 2/3 \ int \ (sec^2u)/sqrt(4sec^2u) \ du`
`\ \ = 2/3 \ int \ (sec^2u)/(2secu) \ du`
`\ \ = 1/3 \ int \ secu \ du`
`\ \ = 1/3 \ int \ secu \ du`
`\ \ = 1/3 \ ln (secu + tanu)+c`

And again using `1+tan^2A-=sec^2A` we write

`sec^2u = 1+tan^2u`
`" " = 1+(9x^2)/4`
`=> sec u =sqrt(1+(9x^2)/4)`

And restoring the substituting we get:

`I = 1/3 \ ln (sqrt(1+(9x^2)/4) + (3x)/2)+c`

Answer 2

`int dx/sqrt(9x^2+4)`

Start with a simple arithmetic adjustment, as clearly we are looking to clean this up first.

`=1/2int dx/sqrt(9/4x^2+1)`

We then say that: `y^2 = 9/4 x^2 implies y = 3/2x` so `dy = 3/2 dx`.

`implies 1/3int dy/sqrt(y^2+1)`

Now we note that: `cosh^2 z - sinh^2 z = 1` so we can sub `y = sinh z` which means that: `dy = cosh z \ dz`

`implies 1/3int (cosh z \ dz)/sqrt(sinh^2 z+1)`

`= 1/3int dz`

`= 1/3 ( z + C)`

Now to reverse the sub's:

`= 1/3 sinh^(-1) y + C`, where C is arbitrary.

`= 1/3 sinh^(-1) (3/2 x) + C`

Now: `sinh^(-1) alpha= ln(alpha+sqrt (alpha^{2}+1})`

`implies 1/3 ( ln(3/2 x+sqrt ((3/2 x)^{2}+1}) )+ C`