# How do you use the comparison test to test for convergence if #1 / (xcosx) dx# from 0 to #pi/2#?

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To use the comparison test for the convergence of the integral (\int_0^{\pi/2} \frac{1}{x\cos(x)} , dx), we need to find a function (f(x)) that is greater than or equal to (\frac{1}{x\cos(x)}) for all (x) in the interval ((0, \pi/2]) and whose integral is known to converge or diverge.

Let's consider (f(x) = \frac{1}{x}). For (x) in the interval ((0, \pi/2]), (x) is positive, and (\cos(x)) is also positive, so (x\cos(x) > 0). Hence, (f(x) = \frac{1}{x} \geq \frac{1}{x\cos(x)}) for all (x) in ((0, \pi/2]).

Now, we know that (\int_1^{\infty} \frac{1}{x} , dx) diverges (harmonic series), so by the comparison test, since (\frac{1}{x} \geq \frac{1}{x\cos(x)}) for all (x) in ((0, \pi/2]), and (\int_1^{\infty} \frac{1}{x} , dx) diverges, then (\int_0^{\pi/2} \frac{1}{x\cos(x)} , dx) also diverges.

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