$\int 1 / (x + \sqrt{1 - x^{2}})$

By Colin Rowlands on Monday, May 06, 2002 - 11:41 am:

I've tried various substitutions which look promising but an algebraic solution to this integral still elludes me. Please help!

\int 1 / (x + \sqrt{1 - x^{2}})

By David Loeffler on Monday, May 06, 2002 - 12:08 pm:

Try substituting

x = \sin (θ)

. Then it is

$I = \int \cos θ d θ / (\sin θ + \cos θ)$ .

Now this is a fairly well know, and moderately difficult, integral; it is one of the questions in Dr Siklos's book of practice STEP questions (IIRC). You can evaluate it in one of two ways: substitute $t = \tan θ / 2$ and wade through lots of partial fractions; or by a cunning trick - define

$I_{1} = \int \cos θ d θ / (\sin θ + \cos θ)$

$I_{2} = \int \sin θ d θ / (\sin θ + \cos θ)$

Now $I_{1} + I_{2} = \int d θ = θ + C$ ; and $I_{1} = I_{2} = \int (\cos θ - \sin θ) / (\sin θ + \cos θ)$

Now the numerator of the integrand is the derivative of the denominator; so $I_{1} - I_{2} = \log (\sin θ + \cos θ) + D$ .

If you add these two equations together, we get

$I = I_{1} = 1 / 2 θ + 1 / 2 \log (\sin θ + \cos θ) + E$

( $E$ is some new arbitrary constant $= 1 / 2 (C + D)$ .)

So that is your integral: $1 / 2 \sin^{- 1} (x) + 1 / 2 \log (x + \sqrt{1 - x^{2}}) + E$ .

David