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

Beginner2022-04-10Added 14 answers

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

$\int}_{0}^{a}\frac{\sqrt{2}}{\sqrt{{a}^{2}-{x}^{2}}}dx=\sqrt{2}(\mathrm{arcsin}\left(\frac{a}{a}\right)-\mathrm{arcsin}\left(\frac{0}{a}\right))={\frac{\pi}{\sqrt{2}}$

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

$\int}_{0}^{1}\frac{1}{\sqrt{1-{x}^{2}}}dx=\frac{\pi}{2$

Since

Therefore, your integral is asymptotic to

Another approach is, by enforcing the substitution x=au, the integral is transformed to

Taking the limit into the integral, you can easily get the limit of the integrand

Then, one can easily evaluate the integral to get

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