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Question Number 57825 by maxmathsup by imad last updated on 13/Apr/19

calculate ∫_0 ^π ((2xsinx)/(3 +cos(2x)))dx .

$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{2}{xsinx}}{\mathrm{3}\:+{cos}\left(\mathrm{2}{x}\right)}{dx}\:. \\ $$

Commented by maxmathsup by imad last updated on 13/Apr/19

let I =∫_0 ^π ((2x sinx)/(3 +cos(2x))) dx changement x =π −t give I =∫_0 ^π ((2(π−t)sint)/(3+cos(2t)))dt = 2π ∫_0 ^π ((sint)/(3+cos(2t))) −I ⇒2I =2π ∫_0 ^π ((sint)/(3+cos(2t)))dt ⇒I =π ∫_0 ^π ((sint)/(3+cos(2t))) dt but ∫_0 ^π ((sint)/(3+cos(2t)))dt =∫_0 ^π ((sint)/(3+2cos^2 t−1)) dt =∫_0 ^π ((sint dt)/(2(1+cos^2 t))) =_(u =cost) −∫_(−1) ^1 ((−du)/(2(1+u^2 ))) =(1/2)∫_(−1) ^1 (du/(1+u^2 )) =(1/2)[arctanu]_(−1) ^1 =(1/2)((π/2)) =(π/4) ⇒ ★I =(π^2 /4) ★

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{2}{x}\:{sinx}}{\mathrm{3}\:+{cos}\left(\mathrm{2}{x}\right)}\:{dx}\:\:{changement}\:{x}\:=\pi\:−{t}\:{give}\: \\ $$$${I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{2}\left(\pi−{t}\right){sint}}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}{dt}\:=\:\mathrm{2}\pi\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sint}}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}\:−{I}\:\Rightarrow\mathrm{2}{I}\:=\mathrm{2}\pi\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{sint}}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}{dt} \\ $$$$\:\Rightarrow{I}\:=\pi\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{sint}}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}\:{dt}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{sint}}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}{dt}\:\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sint}}{\mathrm{3}+\mathrm{2}{cos}^{\mathrm{2}} {t}−\mathrm{1}}\:{dt}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sint}\:{dt}}{\mathrm{2}\left(\mathrm{1}+{cos}^{\mathrm{2}} {t}\right)} \\ $$$$=_{{u}\:={cost}} \:\:\:\:\:\:−\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\frac{−{du}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[{arctanu}\right]_{−\mathrm{1}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\right)\:=\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$$$\bigstar{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\bigstar \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

∫_0 ^π ((2(π−x)sin(π−x))/(3+cos2(π−x)))dx ∫_0 ^π ((2πsinx−2xsinx)/(3+cos2x))dx 2I=∫_0 ^π ((2πsinx)/(3+cos2x))dx (I/π)=∫_0 ^π ((−d(cosx))/(3+2cos^2 x−1)) ((−I)/π)=(1/2)∫_0 ^π ((d(cosx))/(1+cos^2 x)) ((−I)/π)=(1/2)×∣tan^(−1) (cosx)∣_0 ^π ((−I)/π)=(1/2)×{tan^(−1) (−1)−tan^(−1) (1)} ((−I)/π)=(1/2)×−2×(π/4) I=(π^2 /4)

$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2}\left(\pi−{x}\right){sin}\left(\pi−{x}\right)}{\mathrm{3}+{cos}\mathrm{2}\left(\pi−{x}\right)}{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2}\pi{sinx}−\mathrm{2}{xsinx}}{\mathrm{3}+{cos}\mathrm{2}{x}}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{2}\pi{sinx}}{\mathrm{3}+{cos}\mathrm{2}{x}}{dx} \\ $$$$\frac{{I}}{\pi}=\int_{\mathrm{0}} ^{\pi} \frac{−{d}\left({cosx}\right)}{\mathrm{3}+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}} \\ $$$$\frac{−{I}}{\pi}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{d}\left({cosx}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} {x}} \\ $$$$\frac{−{I}}{\pi}=\frac{\mathrm{1}}{\mathrm{2}}×\mid{tan}^{−\mathrm{1}} \left({cosx}\right)\mid_{\mathrm{0}} ^{\pi} \\ $$$$\frac{−{I}}{\pi}=\frac{\mathrm{1}}{\mathrm{2}}×\left\{{tan}^{−\mathrm{1}} \left(−\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)\right\} \\ $$$$\frac{−{I}}{\pi}=\frac{\mathrm{1}}{\mathrm{2}}×−\mathrm{2}×\frac{\pi}{\mathrm{4}} \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$

Commented by maxmathsup by imad last updated on 13/Apr/19

sir tanmay your answer is correct thanks.

$${sir}\:{tanmay}\:{your}\:{answer}\:{is}\:{correct}\:{thanks}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

most welcome sir

$${most}\:{welcome}\:{sir} \\ $$

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