Integration Questions

Question Number 91892 by john santu last updated on 03/May/20

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{dx}}{{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {x}+{b}^{\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}} {x}}\:? \\$$

Commented by john santu last updated on 03/May/20

$$\int\:\frac{{dx}}{\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\mathrm{sin}\:^{\mathrm{2}} {x}+{a}^{\mathrm{2}} } \\$$$${say}\:{a}^{\mathrm{2}} =\:{p}\:\wedge\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} ={q} \\$$$$\int\:\frac{{dx}}{{q}\:\mathrm{sin}\:^{\mathrm{2}} {x}+{p}}\:= \\$$$${W}\mathrm{ierstrass}\:\mathrm{let}\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{t}\: \\$$$${dx}\:=\:\frac{\mathrm{2t}\:}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt} \\$$$$\mathrm{tobe}\:\mathrm{continue} \\$$

Commented by Prithwish Sen 1 last updated on 03/May/20

$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:\:=\:\frac{\mathrm{1}}{\mathrm{ab}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{btan}\:\mathrm{x}}{\mathrm{a}}\right)\right]_{\mathrm{0}} ^{\pi} \\$$

Commented by mathmax by abdo last updated on 03/May/20

$${I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{{a}^{\mathrm{2}} \:{cos}^{\mathrm{2}} {x}\:+{b}^{\mathrm{2}} \:{sin}^{\mathrm{2}} {x}}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{2}{dx}}{{a}^{\mathrm{2}} \left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right)+{b}^{\mathrm{2}} \left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)} \\$$$$=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{2}{dx}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){cos}\left(\mathrm{2}{x}\right)}\:=_{\mathrm{2}{x}={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){cost}} \\$$$$=_{{e}^{{it}} \:={z}} \:\:\:\:\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\:\:\frac{{dz}}{{iz}\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)×\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\right)} \\$$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{2}{dz}}{{iz}\left\{\mathrm{2}\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({z}+{z}^{−\mathrm{1}} \right)\right\}} \\$$$$=\int\:\:\frac{−\mathrm{2}{i}\:{dz}}{\mathrm{2}\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right){z}\:+\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){z}^{\mathrm{2}} \:+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:{let}\: \\$$$$\varphi\left({z}\right)\:=\frac{−\mathrm{2}{i}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){z}^{\mathrm{2}} \:+\mathrm{2}\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right){z}\:+{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\:{poles}\:{of}\:\varphi? \\$$$$\Delta^{'} \:=\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \:={a}^{\mathrm{4}} \:+{b}^{\mathrm{4}} \:+\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{a}^{\mathrm{4}} −{b}^{\mathrm{4}} \:+\mathrm{2}{a}^{\mathrm{2}} \:{b}^{\mathrm{2}} \\$$$$=\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:\Rightarrow\:{z}_{\mathrm{1}} =\frac{−\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)+\mathrm{2}\mid{ab}\mid}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }=−\frac{\left(\mid{a}\mid−\mid{b}\mid\right)^{\mathrm{2}} }{\mid{a}\mid^{\mathrm{2}} −\mid{b}\mid^{\mathrm{2}} }=−\frac{\mid{a}\mid−\mid{b}\mid}{\mid{a}\mid+\mid{b}\mid}\:\:\:\:\left({we}\:{suppose}\:{a}\neq{b}\right) \\$$$${z}_{\mathrm{2}} =\frac{−\left({a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)−\mathrm{2}\mid{ab}\mid}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:=−\frac{\mid{a}\mid+\mid{b}\mid}{\mid{a}\mid−\mid{b}\mid}\:=\frac{\mathrm{1}}{{z}_{\mathrm{1}} } \\$$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\frac{\mid\mid{a}\mid−\mid{b}\mid\mid}{\mid{a}\mid+\mid{b}\mid}−\mathrm{1}\:<\mathrm{0}\:{because}\:\mid....\mid\leqslant\mid{a}\mid\:+\mid{b}\mid\:\Rightarrow\mid{z}_{\mathrm{2}} \mid−\mathrm{1}>\mathrm{0}\:\Rightarrow \\$$$$\left({z}_{\mathrm{2}} {is}\:{out}\:{of}\:{circle}\right) \\$$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right) \\$$$$\varphi\left({z}\right)\:=\frac{−\mathrm{2}{i}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\frac{−\mathrm{2}{i}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)} \\$$$$=\frac{−\mathrm{2}{i}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)×\frac{\mathrm{4}\mid{ab}\mid}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:=\frac{−{i}}{\mathrm{2}\mid{ab}\mid}\:\Rightarrow\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right)\:{dz}\:=\frac{\mathrm{2}{i}\pi×\left(−{i}\right)}{\mathrm{2}\mid{ab}\mid} \\$$$$=\frac{\pi}{\mid{ab}\mid}\:\Rightarrow\:{I}\:=\frac{\pi}{\mid{a}\mid×\mid{b}\mid}\:\:\:\:\:\:\left({ab}\neq\mathrm{0}\right) \\$$$${if}\:{a}={b}\:\neq\mathrm{0}\:\Rightarrow\:{I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{{a}^{\mathrm{2}} \:}\:=\frac{\pi}{{a}^{\mathrm{2}} } \\$$

Commented by john santu last updated on 04/May/20

$$\mathrm{it}\:\mathrm{does}\:\mathrm{the}\:\mathrm{result}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{zero}? \\$$

Commented by john santu last updated on 04/May/20

$$\mathrm{waw}\:\mathrm{via}\:\mathrm{complex}\:\mathrm{method} \\$$