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Integration Questions

Question Number 162575 by Ar Brandon last updated on 30/Dec/21

$$\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{sin}{x}}{dx} \\$$

Answered by phanphuoc last updated on 30/Dec/21

$${you}\:{can}\:{put}\:{x}={pi}−{t} \\$$

Answered by mindispower last updated on 30/Dec/21

$${du}=\frac{\mathrm{1}}{\mathrm{1}+{sin}\left({x}\right)}\Rightarrow{u}=\frac{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)+{cos}\left(\frac{{x}}{\mathrm{2}}\right)} \\$$ $${IBP}\Rightarrow\mathrm{2}\pi^{\mathrm{2}} −\mathrm{4}\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{{cos}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{sin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\mathrm{xdx} \\$$ $$\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{{cos}\left(\frac{{x}}{\mathrm{2}}\right)+{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{xdx}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left({y}\right)}{{sin}\left({y}\right)+{cos}\left({y}\right)}{ydy} \\$$ $$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ydy}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({y}\right)−{sin}\left({y}\right)}{{sin}\left({y}\right)+{cos}\left({y}\right)}{ydy} \\$$ $$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({y}\right)+{cos}\left({y}\right)\right){dy} \\$$ $$\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\sqrt{\mathrm{2}}\right)+{ln}\left({sin}\left({y}+\frac{\pi}{\mathrm{4}}\right)\right){dy} \\$$ $$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\pi{ln}\left(\sqrt{\mathrm{2}}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left({y}+\frac{\pi}{\mathrm{4}}\right)\right){dy}+\mathrm{2}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({y}+\frac{\pi}{\mathrm{4}}\right)\right){dy} \\$$ $$=\frac{\pi}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}+{ln}\left(\mathrm{2}\right)\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({y}\right)\right){dy}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({y}\right)\right){dy} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({x}\right)\right){dx}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{G}−\boldsymbol{\pi}{ln}\left(\mathrm{2}\right)\right) \\$$ $$\mathrm{2}{G}−\pi\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\$$ $$\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} {dx}}{\mathrm{1}+{sin}\left({x}\right)}=\mathrm{2}\pi^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}{G}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right) \\$$ $$=\pi^{\mathrm{2}} +\mathrm{2}\pi{ln}\left(\mathrm{2}\right)−\mathrm{8}{G} \\$$

Commented byAr Brandon last updated on 30/Dec/21

$$\mathrm{Cool}\:!\:\mathrm{Merci},\:\mathrm{grand}. \\$$

Commented bymindispower last updated on 31/Dec/21

$${Avec}\:{plaisir}\:{Bonne}\:{journee} \\$$

Commented byAr Brandon last updated on 31/Dec/21

$$\mathrm{Meilleure}\:\grave {\mathrm{a}}\:\mathrm{vous}\:! \\$$

Commented bymindispower last updated on 31/Dec/21

$${tu}\:{vas}\:{faire}\:{MP}\:{maths}\:{spe}? \\$$

Commented byAr Brandon last updated on 31/Dec/21

$$\mathrm{Non},\:\mathrm{je}\:\mathrm{n}'\mathrm{ai}\:\mathrm{pas}\:\mathrm{penser}\:\grave {\mathrm{a}}\:\mathrm{le}\:\mathrm{faire}. \\$$ $$\mathrm{Je}\:\mathrm{suis}\:\mathrm{en}\:\mathrm{1}^{\mathrm{er}} \:\mathrm{ann}\acute {\mathrm{e}e}\:\mathrm{IUT}\:\mathrm{actuellement}. \\$$ $$\mathrm{Et}\:\mathrm{vous}\:? \\$$

Commented bymindispower last updated on 31/Dec/21

$${je}\:{suis}\:{en}\:{M}\mathrm{2}\:{Maths}\:{fondamental}\:\left({Topologie}\:{Algebrique}\right) \\$$ $${Bonne}\:{Continuation} \\$$

Commented bymindispower last updated on 01/Jan/22

$${bonjour}\:{je}\:{vais}\:{faire}\:{un}\:{compte} \\$$ $${pour}\:{echanger}\:{pas}\:{de}\:{soucis} \\$$ $$\\$$

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