Evaluating $\int_{x=0}^{1} \frac{\cos^2x}{\sqrt{x}}$ with Taylor series

93 Views Asked by At

I'm asked to use a Taylor series to expand and simplify the integrand, and then look at the leading term behavior near $\infty$ to determine if the integral converges or diverges.

I used the Taylor series for $\cos x$:

$$ \cos x = \sum\limits_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}$$

$$ \cos^2 x = (\cos x)^2 = \left(\frac{x^0}{0!} - \frac{x^2}{2} + H.O.T.\right)^2 = 1 -x^2+\frac{x^4}{24} $$

$$ \frac{\cos^2x}{\sqrt{x}} = \frac{1}{\sqrt{x}} + \mathcal O\left(x^{\frac{3}{2}}\right) $$

Looking at the leading order term, $ \dfrac{1}{\sqrt{x}} $, it appears to converge:

$$ \lim\limits_{x \to 0^+} \frac{1}{\sqrt{x}} = \infty$$

I've gone through the calculations a few times and feel confident about the answer I've outlined, but I'm still getting this one marked wrong. Any mistakes stand out?