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find-0-pi-dx-a-bcosx-2-with-a-gt-b-gt-0-then-give-the-value-of-0-pi-dx-2-cosx-2-

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Question Number 31096 by abdo imad last updated on 02/Mar/18
find ∫_0 ^π (dx/((a+bcosx)^2 )) with a>b>0 then give the value of ∫_0 ^π (dx/((2+cosx)^2 ))
$${find}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{\left({a}+{bcosx}\right)^{\mathrm{2}} }\:{with}\:{a}>{b}>\mathrm{0}\:{then}\:{give}\:{the} \\ $$$${value}\:{of}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\left(\mathrm{2}+{cosx}\right)^{\mathrm{2}} } \\ $$
Commented by abdo imad last updated on 04/Mar/18
let introduce the function f(a)=∫_0 ^π (dx/(a+bcosx)) we have f^′ (a)=−∫_0 ^π (dx/((a+bcosx)^2 )) ⇒∫_0 ^π (dx/((a+bcosx)^2 ))=−f^′ (a) let calculate f(a) ch.tan((x/2))=t give f(a)= ∫_0 ^∞ (1/(a+b ((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^∞ ((2dt)/(a(1+t^2 )+b(1−t^2 ))) =∫_0 ^∞ ((2dt)/(a+b +(a−b)t^2 ))=∫_0 ^∞ ((2dt)/((a+b)(1+((a−b)/(a+b))t^2 ))) then we use the ch.(√((a−b)/(a+b))) t=u⇒ f(a)= (1/(a+b)) ∫_0 ^∞ (2/((1+u^2 ))) (√((a+b)/(a−b))) du = (2/( (√(a^2 −b^2 )))) ∫_0 ^∞ (du/(1+u^2 ))= (π/2) (2/( (√(a^2 −b^2 )))) = (π/( (√(a^2 −b^2 )))) f(a)=π(a^2 −b^2 )^(−(1/2)) ⇒f^′ (a)=((−π)/2)(2a)(a^2 −b^2 )^(−(3/2)) =((−πa)/((a^2 −b^2 )(√(a^2 −b^2 )))) ⇒∫_0 ^π (dx/((a+bcosx)^2 ))= ((πa)/((a^2 −b^2 )(√(a^2 −b^2 )))) 2) let take a=2 and b=1 ⇒ ∫_0 ^π (dx/((2+cosx)^2 ))= ((2π)/((2^2 −1^2 )(√(2^2 −1^2 )))) = ((2π)/(3(√3))) .
$${let}\:{introduce}\:{the}\:{function}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{{a}+{bcosx}}\:{we}\:{have} \\ $$$${f}^{'} \left({a}\right)=−\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{\left({a}+{bcosx}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\left({a}+{bcosx}\right)^{\mathrm{2}} }=−{f}^{'} \left({a}\right)\:{let} \\ $$$${calculate}\:{f}\left({a}\right)\:{ch}.{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{a}+{b}\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dt}}{{a}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+{b}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{dt}}{{a}+{b}\:+\left({a}−{b}\right){t}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dt}}{\left({a}+{b}\right)\left(\mathrm{1}+\frac{{a}−{b}}{{a}+{b}}{t}^{\mathrm{2}} \right)}\:{then} \\ $$$${we}\:{use}\:{the}\:{ch}.\sqrt{\frac{{a}−{b}}{{a}+{b}}}\:{t}={u}\Rightarrow \\ $$$${f}\left({a}\right)=\:\frac{\mathrm{1}}{{a}+{b}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\sqrt{\frac{{a}+{b}}{{a}−{b}}}\:{du} \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} }}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }=\:\frac{\pi}{\mathrm{2}}\:\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}\:=\:\frac{\pi}{\:\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} }} \\ $$$${f}\left({a}\right)=\pi\left({a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow{f}^{'} \left({a}\right)=\frac{−\pi}{\mathrm{2}}\left(\mathrm{2}{a}\right)\left({a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=\frac{−\pi{a}}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} }}\:\Rightarrow\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\left({a}+{bcosx}\right)^{\mathrm{2}} }=\:\frac{\pi{a}}{\left({a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} \right)\sqrt{{a}^{\mathrm{2}} \:−{b}^{\mathrm{2}} }} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{take}\:{a}=\mathrm{2}\:{and}\:{b}=\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{\left(\mathrm{2}+{cosx}\right)^{\mathrm{2}} }=\:\:\frac{\mathrm{2}\pi}{\left(\mathrm{2}^{\mathrm{2}} \:−\mathrm{1}^{\mathrm{2}} \right)\sqrt{\mathrm{2}^{\mathrm{2}} \:−\mathrm{1}^{\mathrm{2}} }}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$

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