How do you integrate #int 1/sqrt(4x^2+16x+8) # using trigonometric substitution?
It is in general a good idea to divide any polynomial that we have to work with through by its leading coefficient; this almost always makes it simpler to work with. So:
Complete the square:
Make a substitution to cause the square root to vanish via a trig identity. In this case we'll use
Recalling the identity
which simplifies considerably to become
The integral of
So we get
Transform back to our original variable,
Rearrange a little to obtain a simpler form:
Note that the
If we wished to write this in the same style as the original question formula, we could use the same observation about the integration constant to add
It's arguable whether it's better to write it so that the expression inside the root matches that in the original integral or so that it is a simpler expression. I think I prefer our first answer myself:
It's more elegant.
Subst. back ,