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Question Number 67525 by mathmax by abdo last updated on 28/Aug/19
let a>b>0 calculate ∫_0 ^(2π) (dx/((a+bsinx)^2 ))
$${let}\:{a}>{b}>\mathrm{0}\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\left({a}+{bsinx}\right)^{\mathrm{2}} } \\ $$
Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19
let consider f(a,b)=∫_0 ^(2π) (dx/(a+bsinx)) = (1/b) ∫_0 ^(2π) (( dx)/((a/b) +sinx)) ∫_0 ^(2π) (( dx)/(c+sinx)) =2iπΣ_(∣z_k ∣<1) Res(f(z),z_k ) with f(z)=(1/(iz)) .(1/(c+((z−z^(−1) )/(2i))))=(2/( z^2 +2icz−1))=(2/((z−z_0 )(z−z_1 ))) with z_0 =−i(c+(√(c^2 −1 )) ) and z_1 = −i(c−(√(c^2 −1)) ) we have ∣z_0 ∣>1 and ∣z_1 ∣<1 ∫_0 ^(2π) (dx/(c+sinx))=2iπ (2/(z_1 −z_0 ))=((2π)/( (√(c^2 −1)) )) Now f(a,b)=(1/b).((2π)/( (√(((a/b))^2 −1)) ))=((2π)/( (√(a^2 −b^2 )) )) ((∂f(a,b))/∂a)= ∫_0 ^(2π) (dx/((a+bsinx)^2 )) =2π ((−((2a)/(2(√(a^2 −b^2 )) )))/((a^2 −b^2 )))=((−4πa)/((a^2 −b^2 )^(3/2) ))
$$\:{let}\:\:{consider}\:\:{f}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{{a}+{bsinx}}\:\:\:=\:\frac{\mathrm{1}}{{b}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\:\:{dx}}{\frac{{a}}{{b}}\:+{sinx}}\:\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\:\:{dx}}{{c}+{sinx}}\:=\mathrm{2}{i}\pi\underset{\mid{z}_{{k}} \mid<\mathrm{1}} {\sum}{Res}\left({f}\left({z}\right),{z}_{{k}} \right)\:\:\:\:\:{with}\:\:{f}\left({z}\right)=\frac{\mathrm{1}}{{iz}}\:.\frac{\mathrm{1}}{{c}+\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}}=\frac{\mathrm{2}}{\:{z}^{\mathrm{2}} +\mathrm{2}{icz}−\mathrm{1}}=\frac{\mathrm{2}}{\left({z}−{z}_{\mathrm{0}} \right)\left({z}−{z}_{\mathrm{1}} \right)}\:\:\:\:\:{with}\:{z}_{\mathrm{0}} =−{i}\left({c}+\sqrt{{c}^{\mathrm{2}} −\mathrm{1}\:\:}\:\right)\:{and}\:\:{z}_{\mathrm{1}} =\:−{i}\left({c}−\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}\:\right)\: \\ $$$${we}\:{have}\:\:\mid{z}_{\mathrm{0}} \mid>\mathrm{1}\:\:{and}\:\:\mid{z}_{\mathrm{1}} \mid<\mathrm{1}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{{c}+{sinx}}=\mathrm{2}{i}\pi\:\frac{\mathrm{2}}{{z}_{\mathrm{1}} −{z}_{\mathrm{0}} }=\frac{\mathrm{2}\pi}{\:\sqrt{{c}^{\mathrm{2}} −\mathrm{1}}\:} \\ $$$${Now}\:\:\:{f}\left({a},{b}\right)=\frac{\mathrm{1}}{{b}}.\frac{\mathrm{2}\pi}{\:\sqrt{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{1}}\:}=\frac{\mathrm{2}\pi}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:}\: \\ $$$$\frac{\partial{f}\left({a},{b}\right)}{\partial{a}}=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{\left({a}+{bsinx}\right)^{\mathrm{2}} }\:=\mathrm{2}\pi\:\frac{−\frac{\mathrm{2}{a}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}=\frac{−\mathrm{4}\pi{a}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\: \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Aug/19
let f(a) =∫_0 ^(2π) (dx/(a+bsinx)) we have f^′ (a) =−∫_0 ^(2π) (dx/((a+bsinx)^2 )) ⇒ ∫_0 ^(2π) (dx/((a+bsinx)^2 )) =−f^′ (a) changement e^(ix) =z give f(a)=∫_(∣z∣=1) (dz/(iz{a+b((z−z^(−1) )/(2i))})) =∫_(∣z∣=1) ((2idz)/(iz{2ai +bz−bz^(−1) })) =∫_(∣z∣=1) ((2idz)/(−2az +biz^2 −bi)) =∫_(∣z∣=1) ((2idz)/(2i^2 az +biz^2 −bi)) =∫_(∣z∣=1) ((2dz)/(2iaz +bz^2 −b)) let W(z) =(2/(bz^2 +2iaz −b)) poles of W? Δ^′ =−a^2 +b^2 =−(a^2 −b^2 )=(i(√(a^2 −b^2 )))^2 ⇒ z_1 =((−ia +i(√(a^2 −b^2 )))/b) and z_2 =((−ia−i(√(a^2 −b^2 )))/b) ∣z_1 ∣−1 =((∣−a+(√(a^2 −b^2 ))∣)/b)−1 =(((√(a^2 −b^2 ))−a)/b)−1 =(((√(a^2 −b^2 ))−a−b)/b)<0 ⇒∣z_1 ∣<1 we have z_1 .z_2 =−1 ⇒∣z_2 ∣=(1/(∣z_1 ∣))>1 ∫_(∣z∣=1) W(z)dz =2iπ Res(W,z_1 ) W(z) =(2/(b(z−z_1 )(z−z_2 ))) ⇒Res(W,z_1 ) =(2/(b(z_1 −z_2 ))) =(2/(b2i((√(a^2 −b^2 ))/b))) =(1/(i(√(a^2 −b^2 )))) ⇒ ∫_(∣z∣=1) W(z)dz =2iπ ×(1/(i(√(a^2 −b^2 )))) =((2π)/( (√(a^2 −b^2 )))) =f(a) f(a) =2π(a^2 −b^2 )^(−(1/2)) ⇒f^′ (a) =2π(−(1/2))(a^2 −b^2 )^(−(3/2)) =−π (a^2 −b^2 )^(−(3/2)) ⇒ ∫_0 ^(2π) (dx/((a+bsinx)^2 )) =(π/((a^2 −b^2 )^(3/2) ))
$${let}\:\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{{a}+{bsinx}}\:\:{we}\:{have}\:{f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{\left({a}+{bsinx}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{\left({a}+{bsinx}\right)^{\mathrm{2}} }\:=−{f}^{'} \left({a}\right)\:\:{changement}\:{e}^{{ix}} \:={z}\:{give} \\ $$$${f}\left({a}\right)=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{{dz}}{{iz}\left\{{a}+{b}\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}\right\}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{idz}}{{iz}\left\{\mathrm{2}{ai}\:+{bz}−{bz}^{−\mathrm{1}} \right\}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{idz}}{−\mathrm{2}{az}\:+{biz}^{\mathrm{2}} \:−{bi}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{idz}}{\mathrm{2}{i}^{\mathrm{2}} {az}\:+{biz}^{\mathrm{2}} −{bi}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{dz}}{\mathrm{2}{iaz}\:+{bz}^{\mathrm{2}} −{b}}\:\:{let}\:{W}\left({z}\right)\:=\frac{\mathrm{2}}{{bz}^{\mathrm{2}} \:+\mathrm{2}{iaz}\:−{b}}\:\:{poles}\:{of}\:{W}? \\ $$$$\Delta^{'} =−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:=−\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\left({i}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\frac{−{ia}\:+{i}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}}\:\:{and}\:{z}_{\mathrm{2}} =\frac{−{ia}−{i}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\frac{\mid−{a}+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\mid}{{b}}−\mathrm{1}\:=\frac{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }−{a}}{{b}}−\mathrm{1}\:=\frac{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }−{a}−{b}}{{b}}<\mathrm{0} \\ $$$$\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1}\:\:\:\:{we}\:{have}\:\:{z}_{\mathrm{1}} .{z}_{\mathrm{2}} =−\mathrm{1}\:\Rightarrow\mid{z}_{\mathrm{2}} \mid=\frac{\mathrm{1}}{\mid{z}_{\mathrm{1}} \mid}>\mathrm{1} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{z}_{\mathrm{1}} \right) \\ $$$${W}\left({z}\right)\:=\frac{\mathrm{2}}{{b}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow{Res}\left({W},{z}_{\mathrm{1}} \right)\:=\frac{\mathrm{2}}{{b}\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}\:=\frac{\mathrm{2}}{{b}\mathrm{2}{i}\frac{\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{{b}}} \\ $$$$=\frac{\mathrm{1}}{{i}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:\Rightarrow\:\int_{\mid{z}\mid=\mathrm{1}} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:×\frac{\mathrm{1}}{{i}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:=\frac{\mathrm{2}\pi}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:={f}\left({a}\right) \\ $$$${f}\left({a}\right)\:=\mathrm{2}\pi\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow{f}^{'} \left({a}\right)\:=\mathrm{2}\pi\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=−\pi\:\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{\left({a}+{bsinx}\right)^{\mathrm{2}} }\:=\frac{\pi}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Aug/19
error at final lines f(a) =2π(a^2 −b^2 )^(−(1/2)) ⇒ f^′ (a) =2π(−(1/2))(2a)(a^2 −b^2 )^(−(3/2)) =((−2πa)/((a^2 −b^2 )^(3/2) )) ⇒ ∫_0 ^(2π) (dx/((a +bsinx)^2 )) =((2πa)/((a^2 −b^2 )^(3/2) ))
$${error}\:{at}\:{final}\:{lines}\:\:{f}\left({a}\right)\:=\mathrm{2}\pi\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow \\ $$$${f}^{'} \left({a}\right)\:=\mathrm{2}\pi\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{2}{a}\right)\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:=\frac{−\mathrm{2}\pi{a}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\left({a}\:+{bsinx}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}\pi{a}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$

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