Question Number 221367 by SdC355 last updated on 01/Jun/25
$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\:\frac{\mathrm{1}}{\mathrm{5}−\mathrm{4sin}\left(\theta\right)}\:\mathrm{d}\theta=?? \\ $$$$\left(\mathrm{Complex}\:\mathrm{integral}\:\mathrm{method}\right) \\ $$
Answered by vnm last updated on 03/Jun/25
![∫_0 ^(2π) (dt/(5−4sint))=∫_0 ^(2π) (dt/(5−4((e^(it) −e^(−it) )/(2i))))= ∫_0 ^(2π) (dt/(5+2i(e^(it) −e^(−it) )))=∫_0 ^(2π) ((ie^(it) dt)/(ie^(it) (5+2i(e^(it) −e^(−it) ))))= ∫_0 ^(2π) ((ie^(it) dt)/(5ie^(it) −2(e^(it) )^2 +2))=[z=e^(it) ]=∮_(∣z∣=1) (dz/(5iz−2z^2 +2))= −(1/2)∮_(∣z∣=1) (dz/((z−(i/2))(z−2i)))=−(1/2)∮_(∣z∣=1) (((2i)/3)∙(1/(z−(i/2)))−((2i)/3)∙(1/(z−2i)))dz= [the point z=2i is outside the contour ∣z∣=1] −(i/3)∮_(∣z∣=1) (dz/(z−(i/2)))=−(i/3)∙2πi=((2π)/3).](https://www.tinkutara.com/question/Q221396.png)
$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{\mathrm{5}−\mathrm{4sin}{t}}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{\mathrm{5}−\mathrm{4}\frac{{e}^{{it}} −{e}^{−{it}} }{\mathrm{2}{i}}}= \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{\mathrm{5}+\mathrm{2}{i}\left({e}^{{it}} −{e}^{−{it}} \right)}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{ie}^{{it}} {dt}}{{ie}^{{it}} \left(\mathrm{5}+\mathrm{2}{i}\left({e}^{{it}} −{e}^{−{it}} \right)\right)}= \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{ie}^{{it}} {dt}}{\mathrm{5}{ie}^{{it}} −\mathrm{2}\left({e}^{{it}} \right)^{\mathrm{2}} +\mathrm{2}}=\left[{z}={e}^{{it}} \right]=\oint_{\mid{z}\mid=\mathrm{1}} \frac{{dz}}{\mathrm{5}{iz}−\mathrm{2}{z}^{\mathrm{2}} +\mathrm{2}}= \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\oint_{\mid{z}\mid=\mathrm{1}} \frac{{dz}}{\left({z}−\frac{{i}}{\mathrm{2}}\right)\left({z}−\mathrm{2}{i}\right)}=−\frac{\mathrm{1}}{\mathrm{2}}\oint_{\mid{z}\mid=\mathrm{1}} \left(\frac{\mathrm{2}{i}}{\mathrm{3}}\centerdot\frac{\mathrm{1}}{{z}−\frac{{i}}{\mathrm{2}}}−\frac{\mathrm{2}{i}}{\mathrm{3}}\centerdot\frac{\mathrm{1}}{{z}−\mathrm{2}{i}}\right){dz}= \\ $$$$\left[\mathrm{the}\:\mathrm{point}\:{z}=\mathrm{2}{i}\:\mathrm{is}\:\mathrm{outside}\:\mathrm{the}\:\mathrm{contour}\:\mid{z}\mid=\mathrm{1}\right] \\ $$$$−\frac{{i}}{\mathrm{3}}\oint_{\mid{z}\mid=\mathrm{1}} \frac{{dz}}{{z}−\frac{{i}}{\mathrm{2}}}=−\frac{{i}}{\mathrm{3}}\centerdot\mathrm{2}\pi{i}=\frac{\mathrm{2}\pi}{\mathrm{3}}. \\ $$