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Question Number 217881 by OmoloyeMichael last updated on 23/Mar/25
Answered by vnm last updated on 23/Mar/25
∫_0 ^(π/2) x(((sin x)/(1+cos^2 x))+((cos x)/(1+sin^2 x)))dx=I f(x)=((sin x)/(1+cos^2 x))+((cos x)/(1+sin^2 x)) F(x)=∫f(x)dx=−∫((d(cos x))/(1+cos^2 x))+∫((d(sin x))/(1+sin^2 x))= −tan^(−1) (cos x)+tan^(−1) (sin x) I=xF(x)∣_0 ^(π/2) −∫_0 ^(π/2) (x)′F(x)dx=(π^2 /8)−∫_0 ^(π/2) F(x)dx F(x)=_(0≤x≤(π/2)) −F((π/2)−x) ⇒ ∫_0 ^(π/2) F(x)dx=0 I=(π^2 /8)
$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {x}\left(\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}\right){dx}={I} \\ $$$${f}\left({x}\right)=\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}} \\ $$$${F}\left({x}\right)=\int{f}\left({x}\right){dx}=−\int\frac{{d}\left(\mathrm{cos}\:{x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}}+\int\frac{{d}\left(\mathrm{sin}\:{x}\right)}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}= \\ $$$$−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cos}\:{x}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}\:{x}\right) \\ $$$${I}={xF}\left({x}\right)\mid_{\mathrm{0}} ^{\pi/\mathrm{2}} −\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({x}\right)'{F}\left({x}\right){dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {F}\left({x}\right){dx} \\ $$$${F}\left({x}\right)\underset{\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}} {=}−{F}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\:\Rightarrow\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {F}\left({x}\right){dx}=\mathrm{0} \\ $$$${I}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Answered by mr W last updated on 23/Mar/25
I=∫_0 ^(π/2) x(((sin x)/(1+cos^2 x))+((cos x)/(1+sin^2 x)))dx=I_1 +I_2 I_1 =∫_0 ^(π/2) x(((sin x)/(1+cos^2 x)))dx I_2 =∫_0 ^(π/2) x(((cos x)/(1+sin^2 x)))dx =∫_0 ^(π/2) (x−(π/2)+(π/2))(((cos (x−(π/2)+(π/2)))/(1+sin^2 (x−(π/2)+(π/2)))))d(x−(π/2)) =∫_(−(π/2)) ^0 (t+(π/2))(((cos (t+(π/2)))/(1+sin^2 (t+(π/2)))))dt =∫_0 ^(−(π/2)) (t+(π/2))(((sin t)/(1+cos^2 t)))dt =∫_0 ^(−(π/2)) t(((sin t)/(1+cos^2 t)))dt+(π/2)∫_0 ^(−(π/2)) (((sin t)/(1+cos^2 t)))dt =−I_1 +(π/2)∫_0 ^(π/2) ((sin t)/(1+cos^2 t))dt =−I_1 −(π/2)∫_0 ^(π/2) ((d(cos t))/(1+cos^2 t)) =−I_1 −(π/2)[tan^(−1) (cos t)]_0 ^(π/2) =−I_1 −(π/2)(0−(π/4)) =−I_1 +(π^2 /8) ⇒I=I_1 +I_2 =(π^2 /8)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\left(\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\left(\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}\right){dx} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\left(\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\right){dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({x}−\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\left({x}−\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\left({x}−\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right)}\right){d}\left({x}−\frac{\pi}{\mathrm{2}}\right) \\ $$$$\:\:\:=\int_{−\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \left({t}+\frac{\pi}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\left({t}+\frac{\pi}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:\left({t}+\frac{\pi}{\mathrm{2}}\right)}\right){dt} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{−\frac{\pi}{\mathrm{2}}} \left({t}+\frac{\pi}{\mathrm{2}}\right)\left(\frac{\mathrm{sin}\:{t}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{t}}\right){dt} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{−\frac{\pi}{\mathrm{2}}} {t}\left(\frac{\mathrm{sin}\:{t}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{t}}\right){dt}+\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{−\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{sin}\:{t}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{t}}\right){dt} \\ $$$$\:\:\:=−{I}_{\mathrm{1}} +\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:{t}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{t}}{dt} \\ $$$$\:\:\:=−{I}_{\mathrm{1}} −\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left(\mathrm{cos}\:{t}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{t}} \\ $$$$\:\:\:=−{I}_{\mathrm{1}} −\frac{\pi}{\mathrm{2}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cos}\:{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\:\:\:=−{I}_{\mathrm{1}} −\frac{\pi}{\mathrm{2}}\left(\mathrm{0}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:=−{I}_{\mathrm{1}} +\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\Rightarrow{I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$

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