integration

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$\int \frac{1}{\cos^3 x - \sin^3 x} \ dx = \int \frac{1}{(\cos x - \sin x)(1 + \sin x \cos x)} \ dx$

$= \int \frac{\cos x - \sin x}{(\cos x - \sin x)^2(1 + \sin x \cos x)} \ dx$

So let $t = \cos x + \sin x$

Then

$d t = (\cos x - \sin x) \ dx \\ \\(\cos x - \sin x)^2 = 2-t^2 \\ \\1 + \sin x \cos x = \frac{t^2+1}{2}$

Then the integral is $\int \frac{2} {(t^2+1)(2-t^2)} \ dt = \frac{2}{3} \left[\int \frac{1}{t^2+1} \ dt + \int \frac{1}{2-t^2} \ dt \right]$

We are interested in the first integral and its easy to see that

$A = \boxed {\frac{2}{3}}$

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