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Question Number 37404 by ajfour last updated on 12/Jun/18

∫_0 ^( 2π) (√(a^2 +b^2 −2abcos θ)) dθ with a>b>0 .

$$\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} \sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\mathrm{cos}\:\theta}\:{d}\theta\: \\ $$ $$\:\:\:\:{with}\:\:{a}>{b}>\mathrm{0}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18

∫_0 ^(2Π) (√(a^2 +b^2 −2ab+2ab−2abcosθ)) dθ ∫_0 ^(2Π) (√((a−b)^2 +2ab×2sin^2 (θ/2))) dθ ∫_0 ^(2Π) (√((a−b)^2 +(2(√(ab)) )^2 sin^2 θ))dθ =(1/(2(√(ab)) ))∫_0 ^(2Π) (√(k^2 +sin^2 θ)) dθ {k=(((a−b))/(2(√(ab)) ))} another way... the vslue of cosθ lies between ±1 (√(a^2 +b^2 −2ab))<(√(a^2 +b^2 −2abcosθ)) <(√(a^2 +b^2 +2ab)) (a−b)∫_0 ^(2Π) dθ<I<(a+b)∫_0 ^(2Π) dθ (a−b)×2Π<I<(a+b)×2Π

$$\int_{\mathrm{0}} ^{\mathrm{2}\Pi} \sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}+\mathrm{2}{ab}−\mathrm{2}{abcos}\theta}\:{d}\theta \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{2}\Pi} \sqrt{\left({a}−{b}\right)^{\mathrm{2}} +\mathrm{2}{ab}×\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}\:{d}\theta \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{2}\Pi} \sqrt{\left({a}−{b}\right)^{\mathrm{2}} +\left(\mathrm{2}\sqrt{{ab}}\:\right)^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{d}\theta \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{ab}}\:}\int_{\mathrm{0}} ^{\mathrm{2}\Pi} \sqrt{{k}^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta}\:{d}\theta\:\:\left\{{k}=\frac{\left({a}−{b}\right)}{\mathrm{2}\sqrt{{ab}}\:}\right\} \\ $$ $${another}\:{way}... \\ $$ $${the}\:{vslue}\:{of}\:{cos}\theta\:{lies}\:{between}\:\pm\mathrm{1} \\ $$ $$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}}<\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{abcos}\theta}\:\:<\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}} \\ $$ $$\left({a}−{b}\right)\int_{\mathrm{0}} ^{\mathrm{2}\Pi} {d}\theta<{I}<\left({a}+{b}\right)\int_{\mathrm{0}} ^{\mathrm{2}\Pi} {d}\theta \\ $$ $$\left({a}−{b}\right)×\mathrm{2}\Pi<{I}<\left({a}+{b}\right)×\mathrm{2}\Pi \\ $$

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