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Question Number 76716 by peter frank last updated on 29/Dec/19
prove that ∫_0 ^π ((xsin x)/(1+cos^2 x))=(π^2 /4)
$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{x}\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$
Commented by abdomathmax last updated on 30/Dec/19
let I =∫_0 ^π ((xsinx)/(1+cos^2 x))dx changement x=π−t give I =−∫_0 ^π (((π−t)sint)/(1+cos^2 t))(−dt)=π ∫_0 ^π ((sint)/(1+cos^2 t))dt−I ⇒ 2I =π ∫_0 ^π ((sint)/(1+cos^2 t))dt we have ∫_0 ^π ((sint)/(1+cos^2 t))dt =_(tan((t/2))=u) ∫_0 ^∞ (((2u)/(1+u^2 ))/(1+(((1−u^2 )/(1+u^2 )))^2 )) ((2du)/(1+u^2 )) =4 ∫_0 ^∞ (u/((1+u^2 )^2 (1+(((1−u^2 )^2 )/((1+u^2 )^2 )))))du =4 ∫_0 ^∞ ((udu)/((1+u^2 )^2 +(1−u^2 )^2 )) =4 ∫_0 ^∞ ((u du)/(u^4 +2u^2 +1 +u^4 −2u^2 +1)) =4 ∫_0 ^∞ ((udu)/(2u^4 +2)) =2 ∫_0 ^∞ ((udu)/(u^4 +1)) =_(u=z^(1/4) ) 2 ∫_0 ^∞ ((z^(1/4) )/(1+z))×(1/4)z^((1/4)−1) dz =(1/2)∫_0 ^∞ (z^((1/2)−1) /(1+z))dz =(1/2)×(π/(sin((π/2)))) =(π/2) ⇒ 2I =(π^2 /2) ⇒ ★I =(π^2 /4)★
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{xsinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx}\:{changement}\:{x}=\pi−{t}\:{give} \\ $$$${I}\:=−\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{t}\right){sint}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\left(−{dt}\right)=\pi\:\int_{\mathrm{0}} ^{\pi} \frac{{sint}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}{dt}−{I}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\pi\:\int_{\mathrm{0}} ^{\pi} \:\frac{{sint}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}{dt}\:\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\frac{{sint}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}{dt}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+\frac{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)}{du} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{udu}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}\:{du}}{{u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}\:+{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{udu}}{\mathrm{2}{u}^{\mathrm{4}} \:+\mathrm{2}}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{udu}}{{u}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$$=_{{u}={z}^{\frac{\mathrm{1}}{\mathrm{4}}} } \:\:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \frac{{z}^{\frac{\mathrm{1}}{\mathrm{4}}} \:}{\mathrm{1}+{z}}×\frac{\mathrm{1}}{\mathrm{4}}{z}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{z}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{1}+{z}}{dz}\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}\right)}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:\bigstar{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\bigstar \\ $$
Commented by peter frank last updated on 30/Dec/19
thank you
$${thank}\:{you} \\ $$
Commented by Tony Lin last updated on 30/Dec/19
∫_0 ^π ((xsinx)/(1+cos^2 x))dx let π−θ=x, dθ=−dx ∫_0 ^π xf(sinx)dx =∫_π ^0 (π−θ)f[sin(π−θ)](−dθ) =∫_0 ^π (π−θ)f[sinθ]dθ =π∫_0 ^π f(sinθ)dθ−∫_0 ^π θf(sinθ)dθ =π∫_0 ^π f(sinx)dx−∫_0 ^π xf(sinx)dx ⇒∫_0 ^π xf(sinx)dx=(π/2)∫_0 ^π f(sinx)dx ∫_0 ^π ((xsinx)/(1+cos^2 x))dx =(π/2)∫_0 ^π ((sinx)/(1+cos^2 x))dx let cosx=t, dt=−sinxdx (π/2)∫_(−1) ^1 (1/(1+t^2 ))dt =πtan^(−1) t∣_0 ^1 =(π^2 /4)
$$\int_{\mathrm{0}} ^{\pi} \frac{{xsinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$${let}\:\pi−\theta={x},\:{d}\theta=−{dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} {xf}\left({sinx}\right){dx} \\ $$$$=\int_{\pi} ^{\mathrm{0}} \left(\pi−\theta\right){f}\left[{sin}\left(\pi−\theta\right)\right]\left(−{d}\theta\right) \\ $$$$=\int_{\mathrm{0}} ^{\pi} \left(\pi−\theta\right){f}\left[{sin}\theta\right]{d}\theta \\ $$$$=\pi\int_{\mathrm{0}} ^{\pi} {f}\left({sin}\theta\right){d}\theta−\int_{\mathrm{0}} ^{\pi} \theta{f}\left({sin}\theta\right){d}\theta \\ $$$$=\pi\int_{\mathrm{0}} ^{\pi} {f}\left({sinx}\right){dx}−\int_{\mathrm{0}} ^{\pi} {xf}\left({sinx}\right){dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} {xf}\left({sinx}\right){dx}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {f}\left({sinx}\right){dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{xsinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$${let}\:{cosx}={t},\:{dt}=−{sinxdx} \\ $$$$\frac{\pi}{\mathrm{2}}\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\pi{tan}^{−\mathrm{1}} {t}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$
Commented by peter frank last updated on 30/Dec/19
thank you
$${thank}\:{you} \\ $$
Answered by Kunal12588 last updated on 30/Dec/19
I=∫_0 ^π ((x sin x)/(1+cos^2 x))dx ⇒I=∫_0 ^π (((π−x) sin (π−x))/(1+cos^2 (π−x)))dx=∫_0 ^π (((π−x)sin x)/(1+cos^2 x))dx [using ∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx] adding red ones 2I= π∫_0 ^π ((sin x)/(1+cos^2 x))dx (((t=cos x⇒dt=−sin x dx)),((x→0⇒t→1 & x→π⇒t→−1)) ) 2I=−π∫_1 ^(−1) (dt/(1+t^2 ))=π∫_(−1) ^1 (dt/(1+t^2 ))=2π∫_0 ^1 (dt/(1+t^2 )) [using −∫_b ^a f(x)dx=∫_a ^b f(x)dx] [and ∫_(−a) ^( a) f(x) dx = { ((∫_0 ^( 2a) f(x) dx ; if f(x) is even function)),((0 ; if f(x) is odd function)) :}] ⇒I=π[tan^(−1) t]_0 ^1 =π[(π/4)−0]=(π^2 /4) ∴∫_0 ^π ((x sin x)/(1+cos^2 x))dx=(π^2 /4) QED
$${I}=\int_{\mathrm{0}} ^{\pi} \frac{{x}\:{sin}\:{x}}{\mathrm{1}+{cos}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\Rightarrow{I}=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right)\:{sin}\:\left(\pi−{x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \:\left(\pi−{x}\right)}{dx}=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right){sin}\:{x}}{\mathrm{1}+{cos}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{using}\:\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx}\right] \\ $$$$\:\:\:\:{adding}\:{red}\:{ones} \\ $$$$\mathrm{2}{I}=\:\pi\int_{\mathrm{0}} ^{\pi} \frac{{sin}\:{x}}{\mathrm{1}+{cos}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{pmatrix}{{t}={cos}\:{x}\Rightarrow{dt}=−{sin}\:{x}\:{dx}}\\{{x}\rightarrow\mathrm{0}\Rightarrow{t}\rightarrow\mathrm{1}\:\&\:{x}\rightarrow\pi\Rightarrow{t}\rightarrow−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{2}{I}=−\pi\int_{\mathrm{1}} ^{−\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\pi\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\mathrm{2}\pi\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{using}\:−\int_{{b}} ^{{a}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\right] \\ $$$$\left[{and}\:\:\underset{−{a}} {\overset{\:\:\:\:\:\:\:\:\:{a}} {\int}}{f}\left({x}\right)\:{dx}\:=\:\:\begin{cases}{\int_{\mathrm{0}} ^{\:\mathrm{2}{a}} {f}\left({x}\right)\:{dx}\:;\:{if}\:{f}\left({x}\right)\:{is}\:{even}\:{function}}\\{\mathrm{0}\:;\:{if}\:{f}\left({x}\right)\:{is}\:{odd}\:{function}}\end{cases}\right] \\ $$$$\Rightarrow{I}=\pi\left[{tan}^{−\mathrm{1}} {t}\right]_{\mathrm{0}} ^{\mathrm{1}} =\pi\left[\frac{\pi}{\mathrm{4}}−\mathrm{0}\right]=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\therefore\int_{\mathrm{0}} ^{\pi} \frac{{x}\:{sin}\:{x}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\:\:\:\:{QED} \\ $$
Commented by peter frank last updated on 30/Dec/19
thank you
$${thank}\:{you} \\ $$

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