Question Number 32349 by abdo imad last updated on 23/Mar/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{xdx}}{\mathrm{1}+{sinx}}\:. \\ $$
Commented by abdo imad last updated on 24/Mar/18
![the ch.tan((x/2))=t give I =∫_0 ^π ((xdx)/(1+sinx)) =∫_0 ^∞ ((2artant)/(1+((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) =4 ∫_0 ^∞ ((arctant)/(1+t^2 +2t))dt =4 ∫_0 ^∞ ((arctant)/((t+1)^2 ))dt ′by parts we get I =4( [−(1/(t+1)) arctant]_o ^(+∞) −∫_0 ^∞ −(1/(t+1)) (dt/(1+t^2 ))) I =4 ∫_0 ^∞ (dt/((t+1)(t^2 +1))) let decompose f(t) = (1/((t+1)(t^2 +1))) =(a/(t+1)) +((bt +c)/(t^2 +1)) a=lim_(t→−1) (t+1)f(t) =(1/2) lim_(t→+∞) tf(t) =0 =a +b ⇒b=−a=−(1/2) ⇒ f(t) = (1/(2(t+1))) +((((−1)/2)t +c)/(t^2 +1)) f(0) =1 = (1/2) +c ⇒ c=(1/2) ⇒f(t)= (1/(2(t+1))) +(1/2) ((−t +1)/(t^2 +1)) I = 2∫_0 ^∞ ( (1/(t+1)) −((t−1)/(t^2 +1)))dt =2∫_0 ^∞ ( (1/(t+1)) −(1/2) ((2t)/(t^2 +1)))dt +2 ∫_0 ^∞ (dt/(t^2 +1)) I = 2 [ln∣t+1∣−ln(√(t^2 +1)) ]_0 ^∞ +π I = 2 [ln∣ ((t+1)/( (√(t^2 +1))))∣]_0 ^∞ +π =0+π ⇒ I =π .](https://www.tinkutara.com/question/Q32430.png)
$${the}\:{ch}.{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{xdx}}{\mathrm{1}+{sinx}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{artant}}{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctant}}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}{t}}{dt}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctant}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:'{by}\:{parts}\:{we}\:{get} \\ $$$${I}\:=\mathrm{4}\left(\:\:\left[−\frac{\mathrm{1}}{{t}+\mathrm{1}}\:{arctant}\right]_{{o}} ^{+\infty} \:−\int_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{{t}+\mathrm{1}}\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$${I}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:{let}\:{decompose} \\ $$$${f}\left({t}\right)\:=\:\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{bt}\:+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\: \\ $$$${a}={lim}_{{t}\rightarrow−\mathrm{1}} \left({t}+\mathrm{1}\right){f}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${lim}_{{t}\rightarrow+\infty} {tf}\left({t}\right)\:=\mathrm{0}\:={a}\:+{b}\:\Rightarrow{b}=−{a}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\:+\frac{\frac{−\mathrm{1}}{\mathrm{2}}{t}\:+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+{c}\:\Rightarrow\:{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{f}\left({t}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\frac{−{t}\:+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${I}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left(\:\:\frac{\mathrm{1}}{{t}+\mathrm{1}}\:−\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left(\:\frac{\mathrm{1}}{{t}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\right){dt}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${I}\:=\:\mathrm{2}\:\left[{ln}\mid{t}+\mathrm{1}\mid−{ln}\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\right]_{\mathrm{0}} ^{\infty} \:\:+\pi \\ $$$${I}\:=\:\mathrm{2}\:\left[{ln}\mid\:\frac{{t}+\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\mid\right]_{\mathrm{0}} ^{\infty} \:+\pi\:=\mathrm{0}+\pi\:\Rightarrow\:{I}\:=\pi\:. \\ $$