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Question Number 46850 by maxmathsup by imad last updated on 01/Nov/18
let a^2 >b^(2 ) +c^2 calculate ∫_0 ^(2π) (dθ/(a+bsinθ +c cosθ))
$${let}\:{a}^{\mathrm{2}} >{b}^{\mathrm{2}\:} +{c}^{\mathrm{2}} \:\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{d}\theta}{{a}+{bsin}\theta\:+{c}\:{cos}\theta} \\ $$
Commented by maxmathsup by imad last updated on 01/Nov/18
residus method let I = ∫_0 ^(2π) (dθ/(a+bsinθ +ccosθ)) changement z =e^(iθ) give I = ∫_(∣z∣=1) (1/(a +b((z−z^(−1) )/(2i))+c((z+z^(−1) )/2))) (dz/(iz)) =∫_(∣z∣=1) (dz/(iz{a+b((z−z^(−1) )/(2i))+c((z+z^(−1) )/2)})) = ∫_(∣z∣=1) ((2dz)/(z{2ia +b(z−z^(−1) )+ci(z+z^(−1) ) })) = ∫_(∣z∣=1) ((2dz)/(2iaz +bz^2 −b +ciz^2 +ci)) =∫_(∣z∣=1) ((2dz)/((b+ci)z^2 +2iaz +ci−b)) let ϕ(z) =(2/((b+ci)z^2 +2iaz +ci−b)) .poles of ϕ? Δ^′ =(ia)^2 −(b+ci)(ci−b) =−a^2 −((ci)^2 −b^2 ) =−a^2 +b^2 +c^2 =−(a^2 −(b^2 +c^2 ))<0 ⇒ z_1 =((−ia +i(√(a^2 −b^2 −c^2 )))/(b+ci)) z_2 =((−ia−i(√(a^2 −b^2 −c^2 )))/(b+ci)) ∣z_1 ∣−1 =((∣a−(√(a^2 −b^2 −c^2 ))∣)/( (√(b^2 +c^2 )))) >1 ⇒z_1 is out of circle . ⇒Res(ϕ,z_1 )=0 ∣z_2 ∣−1 =((∣a+(√(a^2 −b^2 −c^2 ))∣ )/( (√(b^2 +c^2 )))) >1 ⇒ z_2 is out of circle ⇒Res(ϕ,z_(2 ) )=0 ∫_0 ^(2π) ϕ(z)dz =0
$${residus}\:{method}\:\:{let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{d}\theta}{{a}+{bsin}\theta\:+{ccos}\theta}\:\:{changement}\:{z}\:={e}^{{i}\theta} \:{give} \\ $$$${I}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{a}\:+{b}\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}+{c}\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{{dz}}{{iz}\left\{{a}+{b}\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}+{c}\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\right\}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dz}}{{z}\left\{\mathrm{2}{ia}\:+{b}\left({z}−{z}^{−\mathrm{1}} \right)+{ci}\left({z}+{z}^{−\mathrm{1}} \right)\:\right\}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{dz}}{\mathrm{2}{iaz}\:+{bz}^{\mathrm{2}} −{b}\:+{ciz}^{\mathrm{2}} \:+{ci}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{dz}}{\left({b}+{ci}\right){z}^{\mathrm{2}} \:+\mathrm{2}{iaz}\:+{ci}−{b}} \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{\mathrm{2}}{\left({b}+{ci}\right){z}^{\mathrm{2}} \:+\mathrm{2}{iaz}\:+{ci}−{b}}\:.{poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} \:=\left({ia}\right)^{\mathrm{2}} −\left({b}+{ci}\right)\left({ci}−{b}\right)\:=−{a}^{\mathrm{2}} \:−\left(\left({ci}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$$=−{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:+{c}^{\mathrm{2}} \:=−\left({a}^{\mathrm{2}} −\left({b}^{\mathrm{2}} \:+{c}^{\mathrm{2}} \right)\right)<\mathrm{0}\:\Rightarrow\:{z}_{\mathrm{1}} =\frac{−{ia}\:+{i}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{{b}+{ci}} \\ $$$${z}_{\mathrm{2}} =\frac{−{ia}−{i}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{{b}+{ci}} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\frac{\mid{a}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }\mid}{\:\sqrt{{b}^{\mathrm{2}} \:+{c}^{\mathrm{2}} }}\:>\mathrm{1}\:\Rightarrow{z}_{\mathrm{1}} {is}\:{out}\:{of}\:{circle}\:.\:\Rightarrow{Res}\left(\varphi,{z}_{\mathrm{1}} \right)=\mathrm{0} \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\frac{\mid{a}+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }\mid\:}{\:\sqrt{{b}^{\mathrm{2}} \:+{c}^{\mathrm{2}} }}\:\:>\mathrm{1}\:\Rightarrow\:{z}_{\mathrm{2}} {is}\:{out}\:{of}\:{circle}\:\:\Rightarrow{Res}\left(\varphi,{z}_{\mathrm{2}\:} \right)=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\varphi\left({z}\right){dz}\:=\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18
t=tan(θ/2) dt=sec^2 (θ/2)×(1/2)dθ dθ=((2dt)/(1+t^2 )) ∫((2dt)/((1+t^2 )(a+b×((2t)/(1+t^2 ))+c×((1−t^2 )/(1+t^2 ))))) ∫((2dt)/((a+at^2 +2bt+c−ct^2 ))) 2∫(dt/(t^2 (a−c)+t(2b)+a+c)) (2/(a−c))∫(dt/(t^2 +2.t.(b/(a−c))+((a+c)/(a−c)))) (2/(a−c))∫(dt/((t+(b/(a−c)))^2 +((a+c)/(a−c))−(b^2 /((a−c)^2 )))) (2/(a−c))∫(dt/((t+(b/(a−c)))^2 +((a^2 −c^2 −b^2 )/((a−c)^2 )))) (2/(a−c))×(1/( (√((a^2 −b^2 −c^2 )/((a−c)^2 )))))×tan^(−1) (((t+(b/(a−c)))/( (√((a^2 −b^2 −c2)/((a−c)^2 )))))) =(2/( (√(a^2 −b^2 −c^2 ))))tan^(−1) (((t+(b/(a−c)))/( (√(((a^2 −b^2 −c^2 )/((a−c)^2 )) ))))) when θ→0 t→0 θ→2π t→0 thus value of inyregation zero pls check where i am wrong.. since a^2 >(b^2 +c^2 )
$${t}={tan}\frac{\theta}{\mathrm{2}}\:\:\:{dt}={sec}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}{d}\theta \\ $$$${d}\theta=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\int\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}+{b}×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+{c}×\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$\int\frac{\mathrm{2}{dt}}{\left({a}+{at}^{\mathrm{2}} +\mathrm{2}{bt}+{c}−{ct}^{\mathrm{2}} \right)} \\ $$$$\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} \left({a}−{c}\right)+{t}\left(\mathrm{2}{b}\right)+{a}+{c}} \\ $$$$\frac{\mathrm{2}}{{a}−{c}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}.{t}.\frac{{b}}{{a}−{c}}+\frac{{a}+{c}}{{a}−{c}}} \\ $$$$\frac{\mathrm{2}}{{a}−{c}}\int\frac{{dt}}{\left({t}+\frac{{b}}{{a}−{c}}\right)^{\mathrm{2}} +\frac{{a}+{c}}{{a}−{c}}−\frac{{b}^{\mathrm{2}} }{\left({a}−{c}\right)^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{2}}{{a}−{c}}\int\frac{{dt}}{\left({t}+\frac{{b}}{{a}−{c}}\right)^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\left({a}−{c}\right)^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{2}}{{a}−{c}}×\frac{\mathrm{1}}{\:\sqrt{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\left({a}−{c}\right)^{\mathrm{2}} }}}×{tan}^{−\mathrm{1}} \left(\frac{{t}+\frac{{b}}{{a}−{c}}}{\:\sqrt{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}\mathrm{2}}{\left({a}−{c}\right)^{\mathrm{2}} }}}\right) \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{tan}^{−\mathrm{1}} \left(\frac{{t}+\frac{{b}}{{a}−{c}}}{\:\sqrt{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\left({a}−{c}\right)^{\mathrm{2}} }\:}}\right) \\ $$$${when}\:\theta\rightarrow\mathrm{0}\:\:{t}\rightarrow\mathrm{0} \\ $$$$\theta\rightarrow\mathrm{2}\pi\:\:{t}\rightarrow\mathrm{0} \\ $$$${thus}\:{value}\:{of}\:{inyregation}\:{zero} \\ $$$${pls}\:{check}\:{where}\:{i}\:{am}\:{wrong}.. \\ $$$${since}\:{a}^{\mathrm{2}} >\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$ \\ $$

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