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

∫_0 ^π (x^2 /(1+sinx))dx

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

Answered by phanphuoc last updated on 30/Dec/21

you can put x=pi−t

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

Answered by mindispower last updated on 30/Dec/21

du=(1/(1+sin(x)))⇒u=((2sin((x/2)))/(sin((x/2))+cos((x/2)))) IBP⇒2π^2 −4∫_0 ^π ((sin((x/2)))/(cos((x/2))+sin((x/2))))xdx ∫_0 ^π ((sin((x/2)))/(cos((x/2))+sin((x/2))))xdx=4∫_0 ^(π/2) ((sin(y))/(sin(y)+cos(y)))ydy =2∫_0 ^(π/2) ydy−2∫_0 ^(π/2) ((cos(y)−sin(y))/(sin(y)+cos(y)))ydy =(π^2 /4)+2∫_0 ^(π/2) ln(sin(y)+cos(y))dy (π^2 /4)+2∫_0 ^(π/2) ln((√2))+ln(sin(y+(π/4)))dy =(π^2 /4)+πln((√2))+2∫_0 ^(π/4) ln(sin(y+(π/4)))dy+2∫_(π/4) ^(π/2) ln(sin(y+(π/4)))dy =(π/2)((π/2)+ln(2))+2∫_0 ^(π/4) ln(cos(y))dy+2∫_0 ^(π/4) ln(cos(y))dy ∫_0 ^(π/4) ln(cos(x))dx=(1/4)(2G−𝛑ln(2)) 2G−π((ln(2))/2)+(π^2 /4) ∫_0 ^π ((x^2 dx)/(1+sin(x)))=2π^2 −4(2G−((πln(2))/2)+(π^2 /4)) =π^2 +2πln(2)−8G

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

Commented by Ar Brandon last updated on 30/Dec/21

Cool ! Merci, grand.

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

Commented by mindispower last updated on 31/Dec/21

Avec plaisir Bonne journee

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

Commented by Ar Brandon last updated on 31/Dec/21

Meilleure a^� vous !

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

Commented by mindispower last updated on 31/Dec/21

tu vas faire MP maths spe?

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

Commented by Ar Brandon last updated on 31/Dec/21

Non, je n′ai pas penser a^� le faire. Je suis en 1^(er) anne^� e IUT actuellement. Et vous ?

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

Commented by mindispower last updated on 31/Dec/21

je suis en M2 Maths fondamental (Topologie Algebrique) Bonne Continuation

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

Commented by mindispower last updated on 01/Jan/22

bonjour je vais faire un compte pour echanger pas de soucis

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

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