Without knowing what the integral is, it's impossible to say. But the choice of substitution falls to whichever trigonometric identity most closely resembles what you're given.
Recall the Pythagorean identity:
sin²(x) + cos²(x) = 1 ⇒ cos²(x) = 1 - sin²(x)
Dividing through both sides by cos²(x) gives another version,
tan²(x) + 1 = sec²(x)
The choice of substitution then depends on the presence of the polynomial 1 ± y².
There are also the hyperbolic identities,
cosh²(x) - sinh²(x) = 1 ⇒ cosh²(x) = 1 + sinh²(x)
and
1 - tanh²(x) = sech²(x)
so again we see the polynomial 1 ± y².
Then
[tex]\displaystyle \int f(1 - x^2) \, dx = \int f(\cos^2(z)) \, \cos(z) \, dz[/tex]
where we substitute x = sin(z) and dx = cos(z) dz; and
[tex]\displaystyle \int f(1 + x^2) \, dx = \int f(\sec^2(z)) \, \sec^2(z) \, dz[/tex]
where x = tan(z) and dz = sec²(z) dz.
Some more concrete examples:
• substituting x = tan(z) :
[tex]\displaystyle \int \frac{dx}{1 + x^2} = \int \frac{\sec^2(z)}{1 + \tan^2(z)} \, dz \\\\ = \int \frac{\sec^2(z)}{\sec^2(z)} \, dz \\\\ = \int dz \\\\ = z + C \\\\ \tan^{-1}(x) + C[/tex]
• substituting x = sin(z) :
[tex]\displaystyle \int \frac{dx}{\sqrt{1 - x^2}} = \int \frac{\cos(z)}{\sqrt{1-\sin^2(z)}} \, dz \\\\ = \int \frac{\cos(z)}{\sqrt{\cos^2(z)}} \, dz \\\\ = \int \frac{\cos(z)}{|\cos(z)|} \, dz[/tex]
and the integrand reduces to ±1, depending on the sign of cos(z). If cos(z) > 0 (which happens if 0 ≤ x < 1), then we can continue to end up with
[tex]\displaystyle \int dz = z + C = \sin^{-1}(x) + C[/tex]
and otherwise, if cos(z) < 0,
[tex]\displaystyle -\int dz = -z + C = -\sin^{-1}(x) + C[/tex]
