Question Number 35242 by abdo mathsup 649 cc last updated on 17/May/18
$${find}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{xdx}}{\mathrm{1}+{sinx}} \\ $$
Commented by abdo mathsup 649 cc last updated on 17/May/18
![let put I = ∫_0 ^π ((xdx)/(1+sinx)) .changement tan((x/2))=t give I = ∫_0 ^∞ ((2arctant)/(1+ ((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = 4 ∫_0 ^∞ ((arctan(t))/(1+t^2 +2t))dt = 4 ∫_0 ^∞ ((arctan(t))/((t+1)^2 ))dt and by parts ∫_0 ^∞ ((arctan(t))/((t+1)^2 ))dt = [((−1)/(t+1)) arctant]_0 ^(+∞) −∫_0 ^∞ ((−1)/((t+1)(1+t^2 )))dt = ∫_0 ^∞ (dt/((t+1)(t^2 +1))) =∫_0 ^∞ { (1/(t+1)) −((t−1)/(t^2 +1))}dt =∫_0 ^∞ { (1/(t+1)) −(1/2) ((2t)/(t^2 +1))}dt +∫_0 ^∞ (dt/(t^2 +1)) =[ln(((t+1)/(t^2 +1)))]_0 ^(+∞) +(π/2) =(π/2) ⇒ I = 4.(π/2) ⇒ I =2π .](Q35321.png)
$${let}\:{put}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{xdx}}{\mathrm{1}+{sinx}}\:.{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t} \\ $$$${give}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{\mathrm{2}{arctant}}{\mathrm{1}+\:\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}{t}}{dt}\:=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({t}\right)}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${and}\:{by}\:{parts}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{arctan}\left({t}\right)}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\:\left[\frac{−\mathrm{1}}{{t}+\mathrm{1}}\:{arctant}\right]_{\mathrm{0}} ^{+\infty} \:−\int_{\mathrm{0}} ^{\infty} \:\:\frac{−\mathrm{1}}{\left({t}+\mathrm{1}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left\{\:\frac{\mathrm{1}}{{t}+\mathrm{1}}\:−\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right\}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left\{\:\frac{\mathrm{1}}{{t}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\right\}{dt}\:\:+\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\left[{ln}\left(\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right)\right]_{\mathrm{0}} ^{+\infty} \:\:\:\:+\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}\:=\:\mathrm{4}.\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${I}\:=\mathrm{2}\pi\:. \\ $$
Answered by math1967 last updated on 17/May/18
![2I=∫_(0 ) ^π ((xdx)/(1+sinx)) +∫_0 ^π (((π−x)dx)/(1+sin(π−x))) =π∫_0 ^π (dx/(1+sinx))=π∫_0 ^π (dx/(1+cos((π/2)−x))) =π∫_0 ^π (dx/(2cos^2 ((π/4)−(x/2)))) [1+cosx=2cos^2 (x/2)] =π∫_0 ^π sec^2 ((π/4)−(x/2))dx =(π/2)∫_(0 ) ^π sec^2 ((π/4)−(x/2))dx=(π/2)[((tan((π/4)−(x/2)))/((−1)/2))]_0 ^π =−π{tan(−(π/4))−tan(π/4)}=2π ∴I=π](Q35251.png)
$$\mathrm{2}{I}=\underset{\mathrm{0}\:} {\overset{\pi} {\int}}\frac{{xdx}}{\mathrm{1}+{sinx}}\:+\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{\left(\pi−{x}\right){dx}}{\mathrm{1}+{sin}\left(\pi−{x}\right)}\: \\ $$$$=\pi\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dx}}{\mathrm{1}+{sinx}}=\pi\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dx}}{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)} \\ $$$$=\pi\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)}\:\:\left[\mathrm{1}+{cosx}=\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right] \\ $$$$=\pi\underset{\mathrm{0}} {\overset{\pi} {\int}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\underset{\mathrm{0}\:\:\:} {\overset{\pi} {\int}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right){dx}=\frac{\pi}{\mathrm{2}}\left[\frac{{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)}{\frac{−\mathrm{1}}{\mathrm{2}}}\right]_{\mathrm{0}} ^{\pi} \\ $$$$=−\pi\left\{{tan}\left(−\frac{\pi}{\mathrm{4}}\right)−{tan}\frac{\pi}{\mathrm{4}}\right\}=\mathrm{2}\pi \\ $$$$\therefore{I}=\pi \\ $$