# $\lim_{a \to 0^{+}} \int_{0}^{a} \frac{1}{\sqrt{\cos(x)-\cos(a)}} \;dx=\frac{\pi}{\sqrt{2}}$

How can I prove that

$$\lim_{a \to 0^{+}} \int_{0}^{a} \frac{1}{\sqrt{\cos(x)-\cos(a)}} \;dx=\frac{\pi}{\sqrt{2}}$$

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You can expand $\cos(x)$ into a Taylor series, keeping only the first two terms, since all the rest go to zero. You get the integral: $$\int_0^b{\frac{\sqrt{2}}{\sqrt{a^2 - x^2}}}dx = \sqrt{2} \arctan\left(\frac{b}{\sqrt{a^2-b^2}}\right)$$ Since $\lim_{x\rightarrow \infty}{\arctan {x}} = \pi/2$, the result follows.

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Nice solution! (+1) –  Chris's sis the artist Aug 29 '12 at 11:17

Exploiting the trigonometric identity $\cos(x) = 1-2\sin^2(\frac{x}{2})$ and making the change of variables $x=ay$ results in the following integral,

$$a\int _{0}^{1}\!{\frac {1}{\sqrt {-2\, \sin^2 \left( \frac{a}{2} \right) + 2\, \sin^2 \left( \frac{a y}{2} \right) }}}{dy}$$

Using the fact that $\sin(a)\approx a$ when $a \rightarrow 0$ in the above integral yields

$$\sqrt {2}\int _{0}^{1}\!{\frac {1}{\sqrt {1-{y}^{2}}}}{dx} = \frac{\pi}{\sqrt{2}} \,.$$

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There is no need to use approximation. After changing $x=a y$: $$\lim_{a \to 0^+} \int_0^a \frac{dx}{\sqrt{\cos(x)-\cos(a)}} = \int_0^1 \sqrt{\lim_{a \to 0^+} \frac{a^2}{\cos(a y)-\cos(a)}} \mathrm{d} y$$ Evaluating the limit by applying l'Hospital's rule twice, the answer follows. –  Sasha Aug 29 '12 at 12:44
Sasha & Chris: Nice finish. –  Mhenni Benghorbal Aug 29 '12 at 16:11