Question Number 37587 by prof Abdo imad last updated on 15/Jun/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}\:+\sqrt{\mathrm{3}}\:{sin}^{\mathrm{2}} {x}} \\ $$
Commented by math khazana by abdo last updated on 16/Jun/18
$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dx}}{\mathrm{2}\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}+\sqrt{\mathrm{3}}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dx}}{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}\:+\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\:+\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{3}}{cos}\left(\mathrm{2}{x}\right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}+\sqrt{\mathrm{3}}\:\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cos}\left(\mathrm{2}{x}\right)} \\ $$$$=_{\mathrm{2}{x}={t}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}\:+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}}\:+\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}+\sqrt{\mathrm{3}}\:\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}} \\ $$$$={K}\:+{H} \\ $$$${chang}.\:{e}^{{it}} \:={z}\:{give} \\ $$$${K}\:=\:\int_{\mid{z}\mid\:=\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\frac{\mathrm{2}{dz}}{{iz}\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left({z}+{z}^{−\mathrm{1}} \right)\right.} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} \:+\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right){z}\:+\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${let}\:\varphi\left({z}\right)=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} \:+\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right){z}\:+\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} \:=\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{4}\:+\mathrm{4}\sqrt{\mathrm{3}}\:+\mathrm{3}−\mathrm{4}\:+\mathrm{4}\sqrt{\mathrm{3}}\:−\mathrm{3} \\ $$$$=\mathrm{8}\sqrt{\mathrm{3}} \\ $$$${z}_{\mathrm{1}} =\:\frac{−\mathrm{2}−\sqrt{\mathrm{3}}\:\:\:+\mathrm{2}\sqrt{\mathrm{2}}\sqrt{\sqrt{\mathrm{3}}}}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${z}_{\mathrm{2}} =\:\frac{−\mathrm{2}−\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{2}}\:\sqrt{\sqrt{\mathrm{3}}}}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}\:=\frac{\mid\:\mathrm{2}+\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{2}}\sqrt{\sqrt{\mathrm{3}}}\mid}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:−\mathrm{8}\sqrt{\mathrm{3}}\:=\:\mathrm{7}\:−\mathrm{4}\sqrt{\mathrm{3}}\:−\mathrm{8}\sqrt{\mathrm{3}}=\mathrm{7}−\mathrm{12}\sqrt{\mathrm{3}}<\mathrm{0} \\ $$$$\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1}\:{and}\:{we}\:{verify}\:{that}\:\:\mid{z}_{\mathrm{2}} \mid>\mathrm{1} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left({z}_{\mathrm{1}} \:−{z}_{\mathrm{2}} \right)} \\ $$$$=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\frac{\mathrm{4}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}{\mathrm{2}−\sqrt{\mathrm{3}}}}\:=\:\:\frac{−{i}}{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:.\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi\:\frac{−{i}}{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:=\:\frac{\pi}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:\:={K}\:. \\ $$
Commented by math khazana by abdo last updated on 16/Jun/18
$${H}\:=\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\:\:\:\frac{{dt}}{\mathrm{2}+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}} \\ $$$$=_{{t}\:=\mathrm{2}\pi\:+{x}} \:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\:\:\:\frac{{dx}}{\mathrm{2}+\sqrt{\mathrm{3}\:}+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cosx}}\:\:\:={K} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:\Rightarrow \\ $$$${I}\:={H}\:+{K}\:=\:\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\:\sqrt{\sqrt{\mathrm{3}}}}\:\: \\ $$$${I}\:\:=\frac{\pi\sqrt{\mathrm{2}}}{\:\sqrt{\sqrt{\mathrm{3}}}}\:. \\ $$
Answered by ajfour last updated on 15/Jun/18
$$\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} \frac{\mathrm{sec}\:^{\mathrm{2}} {xdx}}{\mathrm{2}+\sqrt{\mathrm{3}}\mathrm{tan}\:^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\:\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}} \\ $$$$=\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}×\frac{\mathrm{3}^{\mathrm{1}/\mathrm{4}} }{\:\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}\right)\:=\pi\:\sqrt{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}}\:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jun/18
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18
$${in}\:{place}\:{of}\:\frac{\Pi}{\mathrm{2}}\:\:\:\:{i}\:{put}\:\frac{\Pi}{\mathrm{4}}\:\:\: \\ $$