Question Number 90997 by jagoll last updated on 27/Apr/20
$$\int\underset{\mathrm{0}} {\overset{\pi} {\:}}\:\frac{\mathrm{sin}\:{x}\:{dx}}{\mathrm{1}+\mathrm{sin}\:{x}} \\ $$
Commented by john santu last updated on 27/Apr/20
Commented by jagoll last updated on 27/Apr/20
$${thank}\:{you} \\ $$
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](https://www.tinkutara.com/question/Q91050.png)
$${I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{sinx}}{\mathrm{1}+{sinx}}{dx}\:\Rightarrow\:{I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}+{sinx}−\mathrm{1}}{\mathrm{1}+{sinx}}{dx} \\ $$$$=\pi\:−\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{1}+{sinx}}\:\:{we}\:{have}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\mathrm{1}+{sinx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$\left.=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}{t}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{2}}{{t}+\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} \:=\mathrm{2}\:\Rightarrow{I}\:=\pi−\mathrm{2} \\ $$