Question Number 35226 by abdo mathsup 649 cc last updated on 16/May/18 | ||
$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\:\frac{{dx}}{{a}\:{sin}^{\mathrm{2}} {x}\:\:+{cos}^{\mathrm{2}} {x}} \\ $$ $${with}\:{a}>\mathrm{0} \\ $$ $$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{g}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{sin}^{\mathrm{2}} {x}}{\left({a}\:{sin}^{\mathrm{2}} {x}\:+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{dx} \\ $$ | ||
Commented byprof Abdo imad last updated on 19/May/18 | ||
$$\left.\mathrm{1}\right){we}\:{have}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dx}}{{a}\:{sin}^{\mathrm{2}} {x}\:+{cos}^{\mathrm{2}} {x}} \\ $$ $$=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dx}}{{a}\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}+\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \\ $$ $$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\frac{{dx}}{{a}\:−{a}\:{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)} \\ $$ $$=\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\:\:\frac{{dx}}{{a}+\mathrm{1}\:+\left(\mathrm{1}−{a}\right)\:{cos}\left(\mathrm{2}{x}\right)} \\ $$ $$=_{\mathrm{2}{x}\:={t}} \:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{a}\:+\left(\mathrm{1}−{a}\right){cost}}\:\frac{{dt}}{\mathrm{2}} \\ $$ $$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\:\:\frac{{dt}}{\mathrm{1}+{a}\:+\left(\mathrm{1}−{a}\right){cost}}\:\:{changement} \\ $$ $${e}^{{it}} \:=\:{z}\:\:{give}\: \\ $$ $${I}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{a}\:+\left(\mathrm{1}−{a}\right)\:\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$ $${I}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{dz}}{{iz}\left(\:\mathrm{2}+\mathrm{2}{a}\:+\left(\mathrm{1}−{a}\right)\left({z}+{z}^{−\mathrm{1}} \right)\right.} \\ $$ $$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{−\mathrm{2}{idz}}{\mathrm{2}\left(\mathrm{1}+{a}\right){z}\:\:+\left(\mathrm{1}−{a}\right){z}^{\mathrm{2}} \:\:+\mathrm{1}−{a}} \\ $$ $${let}\:{consider}\:\varphi\left({z}\right)\:=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{1}−{a}\right)^{} {z}^{\mathrm{2}} \:\:+\mathrm{2}\left(\mathrm{1}+{a}\right){z}\:+\mathrm{1}−{a}} \\ $$ $${poles}\:{of}\:\varphi\:? \\ $$ | ||
Commented byprof Abdo imad last updated on 19/May/18 | ||
$$\Delta^{'} \:\:\:=\left(\mathrm{1}+{a}\right)^{\mathrm{2}} \:−\left(\mathrm{1}−{a}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{a}\:+{a}^{\mathrm{2}} \:−\mathrm{1}\:+\mathrm{2}{a}\:−{a}^{\mathrm{2}} \\ $$ $$=\mathrm{4}{a}\:\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{1}−{a}\:+\mathrm{2}\sqrt{{a}}}{\mathrm{1}−{a}}\:=\frac{{a}−\mathrm{2}\sqrt{{a}}\:+\mathrm{1}}{{a}−\mathrm{1}} \\ $$ $$=\:\frac{\left(\sqrt{{a}}\:−\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{a}}\:−\mathrm{1}\right)\left(\sqrt{{a}}\:+\mathrm{1}\right)}\:=\:\frac{\sqrt{{a}}\:−\mathrm{1}}{\sqrt{{a}}\:\:+\mathrm{1}} \\ $$ $${z}_{\mathrm{2}} \:=\:\frac{−\mathrm{1}−{a}\:−\mathrm{2}\sqrt{{a}}}{\mathrm{1}−{a}}\:=\:\frac{\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} }{{a}−\mathrm{1}}\:=\frac{\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{a}}\:−\mathrm{1}\right)\left(\sqrt{{a}}\:+\mathrm{1}\right)} \\ $$ $$=\:\frac{\sqrt{{a}}\:+\mathrm{1}}{\sqrt{{a}}\:−\mathrm{1}} \\ $$ $$\:\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}\:=\:\frac{\sqrt{{a}^{} }−\mathrm{1}}{\sqrt{{a}}+\mathrm{1}}−\mathrm{1}=\:\frac{\sqrt{{a}}\:−\mathrm{1}\:−\sqrt{{a}}\:−\mathrm{1}}{\sqrt{{a}}\:+\mathrm{1}} \\ $$ $$=\:\frac{−\mathrm{2}}{\sqrt{{a}}\:+\mathrm{1}}\:<\mathrm{0}\Rightarrow\:\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$ $$\mid{z}_{\mathrm{2}} \mid\:−\mathrm{1}\:=\:\:\frac{\mathrm{1}}{\mid{z}_{\mathrm{1}} \mid}\:−\mathrm{1}\:=\:\frac{\mathrm{1}−\mid{z}_{\mathrm{1}} \mid}{\mid{z}_{\mathrm{1}} \mid}\:>\mathrm{0}\:\Rightarrow\:\mid{z}_{\mathrm{2}} \mid>\mathrm{1}\left({to}\right. \\ $$ $$\left.{elominate}\:{from}\:{rrsidus}\right) \\ $$ $$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\:{z}_{\mathrm{1}} \right) \\ $$ $${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \left({z}−{z}_{\mathrm{1}} \right)\varphi\left({z}\right) \\ $$ $${but}\:\varphi\left({z}\right)\:=\:\frac{−\mathrm{2}{i}}{\left(\mathrm{1}−{a}\right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$ $${Res}\left(\varphi,{z}_{\mathrm{1}} \right)=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}\:=\:\frac{−\mathrm{2}{i}}{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} \:−\frac{\mathrm{1}}{{z}_{\mathrm{1}} }\right)} \\ $$ $$=\:\:\frac{−\mathrm{2}{i}\:{z}_{\mathrm{1}} }{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:−\mathrm{1}\right)}\: \\ $$ $$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{−\mathrm{2}{iz}_{\mathrm{1}} }{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:−\mathrm{1}\right)} \\ $$ $$=\:\frac{\mathrm{4}\pi{z}_{\mathrm{1}} }{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:−\mathrm{1}\right)} \\ $$ | ||
Commented byabdo mathsup 649 cc last updated on 20/May/18 | ||
$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\:\frac{\mathrm{4}\pi}{\mathrm{1}−{a}}\:\:\frac{\frac{\sqrt{{a}}\:−\mathrm{1}}{\sqrt{{a}}\:+\mathrm{1}}}{\frac{\left(\sqrt{{a}}−\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} }\:−\mathrm{1}} \\ $$ $$=\:\frac{\mathrm{4}\pi}{\mathrm{1}−{a}}\:\:\frac{\sqrt{{a}}\:−\mathrm{1}}{\sqrt{{a}}\:+\mathrm{1}}\:\frac{\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{a}}−\mathrm{1}\right)^{\mathrm{2}} \:−\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $$=\:\frac{\mathrm{4}\pi\left({a}−\mathrm{1}\right)}{\left(\mathrm{1}−{a}\right)\left(\:{a}\:−\mathrm{2}\sqrt{{a}}\:+\mathrm{1}\:−{a}−\mathrm{2}\sqrt{{a}}\:−\mathrm{1}\right)} \\ $$ $$=\:\frac{\mathrm{4}\pi}{\mathrm{4}\sqrt{{a}}}\:=\:\frac{\pi}{\sqrt{{a}}}\:\:{so} \\ $$ $${f}\left({a}\right)\:=\:\frac{\pi}{\sqrt{{a}}}\:. \\ $$ | ||
Commented byabdo mathsup 649 cc last updated on 20/May/18 | ||
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\partial}{\partial{a}}\left\{\:\:\:\frac{\mathrm{1}}{{asin}^{\mathrm{2}} {x}\:+{cos}^{\mathrm{2}} {x}}\right\}{dx} \\ $$ $$=\:−\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{sin}^{\mathrm{2}} {x}}{\left({a}\:{sin}^{\mathrm{2}} {x}\:+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{dx}\:=−{g}\left({a}\right)\:\Rightarrow \\ $$ $${g}\left({a}\right)\:=−{f}^{'} \left({a}\right)\:\:{but}\:\:\:{f}\left({a}\right)=\:\frac{\pi}{\sqrt{{a}}}\:\Rightarrow\:{f}^{'} \left({a}\right)=\:−\pi\:\frac{\left(\sqrt{{a}}\right)^{'} }{{a}} \\ $$ $$=−\pi\:\:\frac{\mathrm{1}}{\mathrm{2}{a}\sqrt{{a}}}\:\Rightarrow\:\:{g}\left({a}\right)\:=\:\frac{−\pi}{\mathrm{2}{a}\sqrt{{a}}}\:\:. \\ $$ | ||