"trigonometric limit with integral: limα→0∫0αdxcos⁡x−cos⁡α" was answered by our experts

Ormezzani6cuu

Answered question

2022-04-09

trigonometric limit with integral:

lim_{α \to 0} \int_{0}^{α} \frac{d x}{\sqrt{\cos x - \cos α}}

star04iks7

Beginner2022-04-10Added 14 answers

By Tayler expansion,

\cos x a s y \mp 1 - \frac{x^{2}}{2}

Since

α

is small, you can replace the integrand with

\sqrt{\frac{2}{α^{2} - x^{2}}}

\cos x a s y \mp 1 - \frac{x^{2}}{2}

\cos a a s y \mp 1 - \frac{a^{2}}{2}

\cos x - \cos a a s y \mp \frac{12}{a^{2} - x^{2}}

Therefore, your integral is asymptotic to

\int_{0}^{a} \frac{\sqrt{2}}{\sqrt{a^{2} - x^{2}}} d x = \sqrt{2} (\arcsin (\frac{a}{a}) - \arcsin (\frac{0}{a})) = \frac{π}{\sqrt{2}}

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

\int_{0}^{1} \frac{a}{\sqrt{\cos a u - \cos a}} d u

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{π}{\sqrt{2}}

by recalling the famous integral

\int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} d x = \frac{π}{2}

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