Ormezzani6cuu

2022-04-09

trigonometric limit with integral: $\underset{\alpha \to 0}{lim}{\int }_{0}^{\alpha }\frac{dx}{\sqrt{\mathrm{cos}x-\mathrm{cos}\alpha }}$

star04iks7

By Tayler expansion,
$\mathrm{cos}xasy\mp 1-\frac{{x}^{2}}{2}$
Since $\alpha$ is small, you can replace the integrand with $\sqrt{\frac{2}{{\alpha }^{2}-{x}^{2}}}$
$\mathrm{cos}xasy\mp 1-\frac{{x}^{2}}{2}$
$\mathrm{cos}aasy\mp 1-\frac{{a}^{2}}{2}$
$\mathrm{cos}x-\mathrm{cos}aasy\mp \frac{12}{{a}^{2}-{x}^{2}}$
Therefore, your integral is asymptotic to
${\int }_{0}^{a}\frac{\sqrt{2}}{\sqrt{{a}^{2}-{x}^{2}}}dx=\sqrt{2}\left(\mathrm{arcsin}\left(\frac{a}{a}\right)-\mathrm{arcsin}\left(\frac{0}{a}\right)\right)=\frac{\pi }{\sqrt{2}}$
Another approach is, by enforcing the substitution x=au, the integral is transformed to
${\int }_{0}^{1}\frac{a}{\sqrt{\mathrm{cos}au-\mathrm{cos}a}}du$
Taking the limit into the integral, you can easily get the limit of the integrand $\frac{1}{\sqrt{\frac{1-{u}^{2}}{2}}}$
Then, one can easily evaluate the integral to get $\frac{\pi }{\sqrt{2}}$ by recalling the famous integral
${\int }_{0}^{1}\frac{1}{\sqrt{1-{x}^{2}}}dx=\frac{\pi }{2}$

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