Integration Questions

Question Number 61465 by arcana last updated on 02/Jun/19

$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{1}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({t}\right)+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({t}\right)}{dt}=\frac{\mathrm{2}\pi}{{ab}}? \\$$

Commented by maxmathsup by imad last updated on 03/Jun/19

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}\:+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}}\:\Rightarrow{A}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\mathrm{2}{dt}}{{a}^{\mathrm{2}} \left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right)+{b}^{\mathrm{2}} \left(\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)\right.} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\mathrm{2}{dt}}{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){cos}\left(\mathrm{2}{t}\right)}\:=_{\mathrm{2}{t}\:={x}} \:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\:\frac{{dx}}{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){cos}\left({x}\right)} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:\:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){cos}\left({x}\right)}\:+\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\:\frac{{dx}}{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){cos}\left({x}\right)}\:={H}+{K} \\$$$${H}\:=_{{e}^{{ix}} ={z}} \:\:\:\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{{dz}}{{iz}\left\{\:{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\right\}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{2}{dz}}{{iz}\left\{\mathrm{2}\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({z}+{z}^{−\mathrm{1}} \right)\right\}} \\$$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dz}}{\mathrm{2}{i}\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right){z}\:+{i}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){z}^{\mathrm{2}} \:+{i}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)} \\$$$$=\int\:\:\frac{−\mathrm{2}{i}\:{dz}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){z}^{\mathrm{2}} +\mathrm{2}\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right){z}\:+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:{let}\:\varphi\left({z}\right)\:=\frac{−\mathrm{2}{i}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){z}^{\mathrm{2}} \:+\mathrm{2}\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right){z}\:+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\$$$$\Delta_{{d}} ^{'} =\:\:=\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:={a}^{\mathrm{4}} \:+\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:+{b}^{\mathrm{4}} −{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{b}^{\mathrm{4}} \:=\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:\:\Rightarrow \\$$$${z}_{\mathrm{1}} =\frac{−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}\mid{ab}\mid}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\:=−\frac{\left(\mid{a}\mid−\mid{b}\mid\right)^{\mathrm{2}} }{\mid{a}\mid^{\mathrm{2}} −\mid{b}\mid^{\mathrm{2}} }\:=−\frac{\mid{a}\mid−\mid{b}\mid}{\mid{a}\mid+\mid{b}\mid}\:\:\:\:\:\left({a}\neq{b}\right) \\$$$${z}_{\mathrm{2}} =\frac{−{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{2}\mid{ab}\mid}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:=−\frac{\left(\mid{a}\mid+\mid{b}\mid\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:=−\frac{\mid{a}\mid+\mid{b}\mid}{\mid{a}\mid−\mid{b}\mid}\:\left(=\frac{\mathrm{1}}{{z}_{\mathrm{1}} }\right)\mid{z} \\$$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\frac{\mid\mid{a}\mid−\mid{b}\mid\mid}{\mid{a}\mid+\mid{b}\mid}\:−\mathrm{1}\:=\:\frac{\mid\mid{a}\mid−\mid{b}\mid\mid−\left(\mid{a}\mid+\mid{b}\mid\right)}{\left(....\right)}<\mathrm{0} \\$$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:>\mathrm{0}\:\:\left({to}\:{eliminate}\:{from}\:{residus}\:\Rightarrow\right. \\$$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi\:,{z}_{\mathrm{1}} \right)\:{we}\:{have}\:\varphi\left({z}\right)\:=\:\frac{−\mathrm{2}{i}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\nLeftrightarrow \\$$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\frac{−\mathrm{2}{i}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}\:=\frac{−\mathrm{2}{i}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\frac{\mid{a}\mid+\mid{b}\mid}{\mid{a}\mid−\mid{b}\mid}−\frac{\mid{a}\mid−\mid{b}\mid}{\mid{a}\mid+\mid{b}\mid}\right)}\:=\frac{−\mathrm{2}{i}}{\left(\mid{a}\mid+\mid{b}\mid\right)^{\mathrm{2}} −\left(\mid{a}\mid−\mid{b}\mid\right)^{\mathrm{2}} } \\$$$$=\frac{−\mathrm{2}{i}}{\mathrm{4}\mid{ab}\mid}\:=−\frac{{i}}{\mathrm{2}\mid{ab}\mid}\:\Rightarrow\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{−{i}}{\mathrm{2}\mid{ab}\mid}\:=\:\frac{\pi}{\mid{ab}\mid}\:={H} \\$$$${K}\:=_{{x}=\mathrm{2}\pi\:+{t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dt}}{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){cost}}\:={H}\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}\:+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}}\:=\mathrm{2}{H}\:\:=\:\frac{\mathrm{2}\pi}{\mid{ab}\mid} \\$$$${if}\:{a}={b}\:{we}\:{get}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {t}\:+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {t}}\:=\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:\left(\mathrm{2}\pi\right)\:=\frac{\mathrm{2}\pi}{{a}^{\mathrm{2}} }\:. \\$$

Answered by tanmay last updated on 03/Jun/19

$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sec}^{\mathrm{2}} {t}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {tan}^{\mathrm{2}} {t}}{dt} \\$$$$\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{d}\left({tant}\right)}{\left(\frac{{a}}{{b}^{} }\right)^{\mathrm{2}} +\left({tant}\right)^{\mathrm{2}} }{dt} \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({t}\right){dt}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({t}\right){dt}\:\:{when}\:{f}\left(\mathrm{2}{a}−{t}\right)={f}\left({t}\right) \\$$$$\frac{\mathrm{2}}{{b}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \frac{{d}\left({tant}\right)}{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\left({tant}\right)^{\mathrm{2}} } \\$$$$=\frac{\mathrm{2}}{{b}^{\mathrm{2}} }×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left({tant}\right)}{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\left({tant}\right)^{\mathrm{2}} } \\$$$$=\frac{\mathrm{4}}{{b}^{\mathrm{2}} }×\frac{\mathrm{1}}{\left(\frac{{a}}{{b}}\right)}×\mid{tan}^{−\mathrm{1}} \left(\frac{{tant}}{\frac{{a}}{{b}}}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$$$=\frac{\mathrm{4}}{{ab}}×\left[{tan}^{−\mathrm{1}} \left(\infty\right)−{tan}^{−\mathrm{1}} \left(\mathrm{0}\right)\right] \\$$$$=\frac{\mathrm{4}}{{ab}}×\frac{\pi}{\mathrm{2}}=\frac{\mathrm{2}\pi}{{ab}} \\$$