Question Number 38210 by prof Abdo imad last updated on 22/Jun/18
$${let}\:{f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{{a}\:+{sin}^{\mathrm{2}} \theta}\:\:\:\left({a}\:{from}\:{R}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{g}\left({a}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{d}\theta}{\mathrm{1}+{sin}^{\mathrm{2}} \theta}\:{and}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{d}\theta}{\mathrm{2}+{sin}^{\mathrm{2}} \theta} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{\left(\mathrm{3}\:+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:. \\ $$
Commented by math khazana by abdo last updated on 23/Jun/18
$${we}\:{have}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{d}\theta}{{a}\:+\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{2}{d}\theta}{\mathrm{2}{a}+\mathrm{1}\:−{cos}\left(\mathrm{2}\theta\right)}\:=_{\mathrm{2}\theta={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{\mathrm{2}}{\mathrm{2}{a}+\mathrm{1}−{cost}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}{a}+\mathrm{1}\:−{cost}}\:\:{changement}\:{e}^{{it}} ={z}\:{give} \\ $$$${f}\left({a}\right)\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{a}+\mathrm{1}−\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−\mathrm{2}{idz}}{{z}\left\{\:\:\mathrm{4}{a}+\mathrm{2}−{z}−{z}^{−\mathrm{1}} \right\}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{\left(\mathrm{4}{a}+\mathrm{2}\right){z}−{z}^{\mathrm{2}} −\mathrm{1}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{idz}}{{z}^{\mathrm{2}} −\left(\mathrm{4}{a}+\mathrm{2}\right){z}\:+\mathrm{1}} \\ $$$${let}\:\varphi\left({z}\right)=\:\frac{\mathrm{2}{i}}{{z}^{\mathrm{2}} \:−\left(\mathrm{4}{a}+\mathrm{2}\right){z}\:+\mathrm{1}}\:.{poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} \:=\left(\mathrm{2}{a}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{1}=\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{4}{a}\:=\mathrm{4}{a}\left({a}+\mathrm{1}\right) \\ $$$${case}\:\mathrm{1}\:\:{a}\left({a}+\mathrm{1}\right)>\mathrm{0}\:\Rightarrow{z}_{\mathrm{1}} =\:\mathrm{2}{a}+\mathrm{1}\:+\mathrm{2}\sqrt{{a}^{\mathrm{2}} \:+{a}} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}{a}+\mathrm{1}−\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{a}} \\ $$$$\mid{z}_{\mathrm{2}} \mid\:−\mathrm{1}\:=\mathrm{2}{a}−\mathrm{2}\sqrt{{a}^{\mathrm{2}} \:+{a}}<\mathrm{0}\:{and}\:{its}\:{clear}\:{that} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}>\mathrm{0}\left({to}\:{elominate}\:{from}\:{residus}\right) \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:=\frac{\mathrm{2}{i}}{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }\:\mathrm{2}{i}\pi \\ $$$$=\frac{−\mathrm{4}\pi}{−\mathrm{4}\sqrt{{a}^{\mathrm{2}} +{a}}}\:=\:\frac{\pi}{\:\sqrt{{a}^{\mathrm{2}} +{a}}}\:\Rightarrow \\ $$$${f}\left({a}\right)=\frac{\pi}{\:\sqrt{{a}^{\mathrm{2}} \:+{a}}} \\ $$$${case}\:\mathrm{2}\:{a}\left({a}+\mathrm{1}\right)<\mathrm{0}\:\:\Leftrightarrow−\mathrm{1}<{a}<\mathrm{0}\:\Rightarrow\Delta^{,} =−\mathrm{4}\left(−{a}\left({a}+\mathrm{1}\right)\right) \\ $$$$=\left(\mathrm{2}{i}\sqrt{−{a}^{\mathrm{2}} −{a}}\right)^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\:\mathrm{2}{a}+\mathrm{1}\:+\mathrm{2}{i}\sqrt{−{a}^{\mathrm{2}} −{a}} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}{a}+\mathrm{1}−\mathrm{2}{i}\sqrt{−{a}^{\mathrm{2}} −{a}} \\ $$$$\mid{z}_{\mathrm{1}} \mid=\sqrt{\:\left(\mathrm{2}{a}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}\left(−{a}^{\mathrm{2}} −{a}\right)} \\ $$$$=\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}\:+\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{a}}=\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid=\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,{z}_{\mathrm{1}} \right)+{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\right\} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\:\frac{\mathrm{2}{i}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} } \\ $$$${Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:=\:\frac{\mathrm{2}{i}}{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }\:\Rightarrow\:\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}=\mathrm{0}\:\Rightarrow{f}\left({a}\right)=\mathrm{0} \\ $$
Commented by math khazana by abdo last updated on 23/Jun/18
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\partial}{\partial{a}}\left(\:\frac{\mathrm{1}}{{a}+{sin}^{\mathrm{2}} \theta}\right){d}\theta \\ $$$$=−\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\frac{{d}\theta}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:=−{g}\left({a}\right)\:\Rightarrow \\ $$$${g}\left({a}\right)\:=−{f}^{'} \left({a}\right)\: \\ $$$${if}\:\:{a}\left({a}+\mathrm{1}\right)<\mathrm{0}\:\Rightarrow{g}\left({a}\right)=\mathrm{0} \\ $$$${if}\:{a}\left({a}+\mathrm{1}\right)>\mathrm{0}\:{f}\left({a}\right)=\:\frac{\pi}{\:\sqrt{{a}^{\mathrm{2}} \:+{a}}}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)\:=\:\pi\left\{\left({a}^{\mathrm{2}} +{a}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right\}^{'} \:=−\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} \:+{a}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=\:\frac{−\pi\left(\mathrm{2}{a}+\mathrm{1}\right)}{\mathrm{2}\left({a}^{\mathrm{2}} +{a}\right)\sqrt{{a}^{\mathrm{2}} \:+{a}}}\:\Rightarrow \\ $$$${g}\left({a}\right)\:=\:\frac{\left(\mathrm{2}{a}+\mathrm{1}\right)\pi}{\mathrm{2}\left({a}^{\mathrm{2}} \:+{a}\right)\sqrt{{a}^{\mathrm{2}} \:+{a}}}\:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 23/Jun/18
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{proved}\:{that}\:\:{f}\left({a}\right)=\frac{\pi}{\:\sqrt{{a}^{\mathrm{2}} \:+{a}}}\:{if}\:{a}\left({a}+\mathrm{1}\right)>\mathrm{0} \\ $$$${so}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{d}\theta}{\mathrm{1}+{sin}^{\mathrm{2}} \theta}\:={f}\left(\mathrm{1}\right)\:=\:\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{d}\theta}{\mathrm{2}+{sin}^{\mathrm{2}} \theta}\:=\:{f}\left(\mathrm{2}\right)=\:\frac{\pi}{\:\sqrt{\mathrm{6}}} \\ $$$$ \\ $$$$\frac{}{} \\ $$
Commented by math khazana by abdo last updated on 23/Jun/18
$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{d}\theta}{\left(\mathrm{3}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:={g}\left(\mathrm{3}\right)=\:\frac{\mathrm{7}\pi}{\mathrm{24}\sqrt{\mathrm{12}}}\:=\frac{\mathrm{7}\pi}{\mathrm{48}\sqrt{\mathrm{3}}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18
$$\int_{\mathrm{0}} ^{\Pi} \frac{{d}\theta}{{a}+{sin}^{\mathrm{2}} \theta} \\ $$$$\int\frac{{sec}^{\mathrm{2}} \theta}{{asec}^{\mathrm{2}} \theta+{tan}^{\mathrm{2}} \theta}{d}\theta \\ $$$${t}={tan}\theta\:\:{dt}={sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\int\frac{{dt}}{{a}+{at}^{\mathrm{2}} +{t}^{\mathrm{2}} } \\ $$$$\int\frac{{dt}}{{a}+{t}^{\mathrm{2}} \left(\mathrm{1}+{a}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}}\int\frac{{dt}}{\frac{{a}}{\mathrm{1}+{a}}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}}×\frac{\mathrm{1}}{\frac{\sqrt{{a}}\:}{\:\sqrt{\mathrm{1}+{a}}}}{tan}^{−\mathrm{1}} \left(\frac{{t}}{\frac{\sqrt{{a}}}{\:\sqrt{\mathrm{1}+{a}}}}\right) \\ $$$$=\mathrm{2}\mid\frac{\mathrm{1}}{\:\sqrt{{a}}\:×\sqrt{\mathrm{1}+{a}}\:}×{tan}^{−\mathrm{1}} \left(\frac{{tan}\theta}{\frac{\sqrt{{a}}\:}{\:\sqrt{\mathrm{1}+{a}}\:}}\right)\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \: \\ $$$$=\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{{a}}\:×\sqrt{\mathrm{1}+{a}}}×\frac{\Pi}{\mathrm{2}} \\ $$$$=\frac{\Pi}{\:\sqrt{{a}+{a}^{\mathrm{2}} }}\:\:\mathscr{A}{ns} \\ $$$$ \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\Pi} {f}\left(\theta\right){d}\theta=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {f}\left(\theta\right){d}\theta\:\:{when}\:{f}\left(\Pi−\theta\right)={f}\left(\theta\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18
$${g}\left({a}\right)=\int_{\mathrm{0}} ^{\Pi} \frac{{d}\theta}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\Pi} \frac{{d}\theta}{{a}+{sin}^{\mathrm{2}} \theta} \\ $$$$\frac{{df}\left({a}\right)}{{da}}=\int_{\mathrm{0}} ^{\Pi} \frac{\partial}{\partial{a}}\frac{\mathrm{1}}{\left({a}+{sin}^{\mathrm{2}} \theta\right)}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\Pi} \frac{−\mathrm{1}}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }{d}\theta \\ $$$${so}\frac{{df}\left({a}\right)}{{da}}=−{g}\left({a}\right) \\ $$$${g}\left({a}\right)=−\frac{{d}}{{da}}\left(\frac{\Pi}{\:\sqrt{{a}+{a}^{\mathrm{2}} }\:}\right) \\ $$$${g}\left({a}\right)=−\Pi×\frac{−\mathrm{1}}{\mathrm{2}\left({a}+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }×\left(\mathrm{1}+\mathrm{2}{a}\right) \\ $$$${g}\left({a}\right)=\frac{\Pi\left(\mathrm{1}+\mathrm{2}{a}\right)}{\mathrm{2}\left({a}+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:\:{ANS} \\ $$$${i}\:{have}\:{proved}\:\int_{\mathrm{0}} ^{\Pi} \frac{{d}\theta}{{a}+{sin}^{\mathrm{2}} \theta}=\frac{\Pi}{\:\sqrt{{a}+{a}^{\mathrm{2}} }} \\ $$$$ \\ $$