Finding the Value of a Trigonometric Integral with Radical

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Discussion Overview

The discussion revolves around evaluating the integral $$\int_0^{2\pi}\sqrt{\dfrac{1-\cos{x}}{2}}\,dx$$. Participants explore various steps in the evaluation process, including transformations and simplifications, while questioning the final result.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related

Main Points Raised

  • Some participants express the integral in a simplified form, noting that $$\cos x = 1 - 2\sin^2\frac x2$$ can be used to transform the integral.
  • One participant rewrites the integral as $$I=\int _0^{2\pi }\sin{\frac x2}\,dx$$ and calculates it, leading to a result of $$I=-2\cos{\dfrac x2}\biggr| _0^{2\pi }$$.
  • Another participant questions the approach, stating they do not see how the evaluation leads to the answer of 4.
  • A later reply calculates the definite integral and arrives at the conclusion that it equals 4, but this is met with skepticism from others regarding the steps taken to reach that conclusion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the evaluation of the integral. While one participant claims the result is 4, others express uncertainty and challenge the reasoning leading to that conclusion.

Contextual Notes

There are unresolved steps in the evaluation process, particularly regarding the transition from the integral of the sine function to the final result. Participants have not fully clarified all assumptions or conditions involved in their calculations.

karush
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Evaluate
$$\int_0^{2\pi}\sqrt{\dfrac{1-\cos{x}}{2}}\,dx$$
ok my baby step is
$$\int _0^{2\pi }\frac{\sqrt{1-\cos \left(x\right)}}{\sqrt{2}}dx$$
then ?

W|A said the answer was 4
 
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karush said:
Evaluate
$$\int_0^{2\pi}\sqrt{\dfrac{1-\cos{x}}{2}}\,dx$$
ok my baby step is
$$\int _0^{2\pi }\frac{\sqrt{1-\cos \left(x\right)}}{\sqrt{2}}dx$$
then ?

W|A said the answer was 4
$\cos x = 1 - 2\sin^2\frac x2$
 
$$I=\int _0^{2\pi }\frac{\sqrt{1-\cos \left(x\right)}}{\sqrt{2}}dx=
\int _0^{2\pi }\dfrac{\sqrt{1-\left(1 - 2\sin^2\dfrac x2\right)}}{\sqrt{2}}dx=
\int _0^{2\pi }\dfrac{\sqrt{2\sin^2{\dfrac x2}}}{\sqrt{2}}=\int _0^{2\pi }\sin{\frac x2}\,dx $$
then
$$I=-2\cos{\dfrac x2}\biggr| _0^{2\pi }=$$

ok I don't see this approaching 4
 
karush said:
$$I=\int _0^{2\pi }\frac{\sqrt{1-\cos \left(x\right)}}{\sqrt{2}}dx=
\int _0^{2\pi }\dfrac{\sqrt{1-\left(1 - 2\sin^2\dfrac x2\right)}}{\sqrt{2}}dx=
\int _0^{2\pi }\dfrac{\sqrt{2\sin^2{\dfrac x2}}}{\sqrt{2}}=\int _0^{2\pi }\sin{\frac x2}\,dx $$
then
$$I=-2\cos{\dfrac x2}\biggr| _0^{2\pi }=$$

ok I don't see this approaching 4
$$I=-2\cos{\dfrac x2}\biggr| _0^{2\pi }= -2(\cos\pi - \cos0) = -2(-1 - 1) = 4$$
 

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