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Question Number 90997 by jagoll last updated on 27/Apr/20

∫ _0 ^π  ((sin x dx)/(1+sin x))

$$\int\underset{\mathrm{0}} {\overset{\pi} {\:}}\:\frac{\mathrm{sin}\:{x}\:{dx}}{\mathrm{1}+\mathrm{sin}\:{x}} \\ $$

Commented by john santu last updated on 27/Apr/20

Commented by jagoll last updated on 27/Apr/20

thank you

$${thank}\:{you} \\ $$

Commented by mathmax by abdo last updated on 27/Apr/20

I =∫_0 ^π  ((sinx)/(1+sinx))dx ⇒ I =∫_0 ^π  ((1+sinx−1)/(1+sinx))dx  =π −∫_0 ^π  (dx/(1+sinx))  we have ∫_0 ^π   (dx/(1+sinx)) =_(tan((x/2))=t)  ∫_0 ^∞   ((2dt)/((1+t^2 )(1+((2t)/(1+t^2 )))))  =∫_0 ^∞   ((2dt)/(1+t^2  +2t)) =∫_0 ^∞   ((2dt)/((t+1)^2 )) =−(2/(t+1))]_0 ^∞  =2 ⇒I =π−2

$${I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{sinx}}{\mathrm{1}+{sinx}}{dx}\:\Rightarrow\:{I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}+{sinx}−\mathrm{1}}{\mathrm{1}+{sinx}}{dx} \\ $$$$=\pi\:−\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{1}+{sinx}}\:\:{we}\:{have}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\mathrm{1}+{sinx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$\left.=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}{t}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{2}}{{t}+\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} \:=\mathrm{2}\:\Rightarrow{I}\:=\pi−\mathrm{2} \\ $$

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