Verified answer

Since cos x = Σ(n = 0 to ∞) (-1)^n x^(2n)/(2n)!, we have

(1 - cos x)/x

= (1/x) [1 - Σ(n = 0 to ∞) (-1)^n x^(2n)/(2n)!]

= (1/x) [1 - (1 + Σ(n = 1 to ∞) (-1)^n x^(2n)/(2n)!)]

= (1/x) * -Σ(n = 1 to ∞) (-1)^n x^(2n)/(2n)!)

= Σ(n = 1 to ∞) (-1)^(n+1) x^(2n-1)/(2n)!.

So, ∫(x = 0 to √(1/3)) (1 - cos x) dx/x

= Σ(n = 1 to ∞) (-1)^(n+1) x^(2n)/((2n) * (2n)!) {for x = 0 to √(1/3)}

= Σ(n = 1 to ∞) (-1)^(n+1) (1/3)^n/((2n) * (2n)!)

= Σ(n = 1 to ∞) (-1)^(n+1) / ((2n) * 3^n * (2n)!).

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To estimate this integral, observe that the answer above is an alternating series.

So, the error after n terms is bounded above by the absolute value of the (n+1)th term:

==> We want 1/((2n+2) * 3^(n+1) * (2n+2)!) < 0.00001

==> We want (2n+2) * 3^(n+1) * (2n+2)! > 100,000

==> n = 2 (or higher) by trial and error.

So, ∫(x = 0 to √(1/3)) (1 - cos x) dx/x

≈ Σ(n = 1 to 2) (-1)^(n+1) / ((2n) * 3^n * (2n)!)

≈ 0.08218 (rounded to five decimal places).

I hope this helps!