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

Try substituting x =
sin θ.  Then it is

I = òcos
θ dθ/(sin θ
+ cos θ).

Now this is a fairly well known, and moderately difficult, integral. 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

I₁ = òcos θ
dθ/(sin θ + cos θ)

I₂ = òsin θ
dθ/(sin θ + cos θ)

Now,

I₁ + I₂ = ò
dθ = θ + C;
and

I₁ - I₂ =
ò(cos θ
- sin θ) dθ / (sin θ +
cos θ)

Now the numerator of the integrand is the derivative of the denominator;
so

I₁ - I₂ = log( sin θ +
cos θ ) + D.

If you add these two equations together, and divide by 2, we get

I = I₁ = ½ θ + ½ log( sin θ
+ cos θ ) + E

(E is some new arbitrary constant = ½(C+D).)

So that is your integral: ½ sin^-1(x) + ½ log (x + sqrt(1-x²) ) + E.