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Question Number 76779 by mathmax by abdo last updated on 30/Dec/19
calculate ∫_0 ^π ((x^2 cosx)/(3+sin^2 x))dx
$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{x}^{\mathrm{2}} {cosx}}{\mathrm{3}+{sin}^{\mathrm{2}} {x}}{dx} \\ $$
Commented by mathmax by abdo last updated on 10/Jan/20
let A =∫_0 ^π ((x^2 cosx)/(3+sin^2 x))dx changement tan((x/2))=t give A =∫_0 ^∞ ((4 arctan^2 (t)×((1−t^2 )/(1+t^2 )))/(3+((4t^2 )/((1+t^2 )^2 ))))×((2dt)/(1+t^2 )) =8 ∫_0 ^∞ (((1−t^2 )arctan^2 t)/((1+t^2 )^2 (3+((4t^2 )/((1+t^2 )^2 )))))dt =8 ∫_0 ^∞ (((1−t^2 )arctan^2 t)/(3(t^2 +1)^2 +4t^2 ))dt =8 ∫_0 ^∞ (((1−t^2 )(arctant)^2 )/(3(t^4 +2t^2 +1)+4t^2 )) =8 ∫_0 ^∞ (((1−t^2 )(arctant)^2 )/(3t^4 +10t^2 +3))dt let solve 3t^4 +10t^2 +3=0 ⇒3u^4 +10u +3=0 →Δ^′ =5^2 −9 =16 ⇒u_1 =((−5+4)/3) =−(1/3) u_2 =((−5−4)/3) =−3 ⇒3t^4 +10t^2 +3 =3(t^2 +(1/3))(t^2 +3) ⇒ A =8 ∫_0 ^∞ (((1−t^2 )(arctant)^2 )/(3(t^2 +(1/3))(t^2 +3)))dt ⇒ (3/4) A =∫_(−∞) ^∞ (((1−t^2 )(arctant)^2 )/((t^2 +(1/3))(t^2 +3)))dt letW(z)=(((1−z^2 )(arctanz)^2 )/((z^2 +(1/3))(z^2 +3))) ⇒ W(z) =(((1−z^2 )(arctanz)^2 )/((z−(i/( (√3))))(z+(i/( (√3))))(z−i(√3))(z+i(√3)))) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,(i/( (√3))))+Res(W,i(√3))} ...be continued...
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{x}^{\mathrm{2}} \:{cosx}}{\mathrm{3}+{sin}^{\mathrm{2}} {x}}{dx}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{4}\:{arctan}^{\mathrm{2}} \left({t}\right)×\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{3}+\frac{\mathrm{4}{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }}×\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right){arctan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{3}+\frac{\mathrm{4}{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)}{dt} \\ $$$$=\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right){arctan}^{\mathrm{2}} {t}}{\mathrm{3}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}{t}^{\mathrm{2}} }{dt}\:=\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left({arctant}\right)^{\mathrm{2}} }{\mathrm{3}\left({t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\right)+\mathrm{4}{t}^{\mathrm{2}} } \\ $$$$=\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left({arctant}\right)^{\mathrm{2}} }{\mathrm{3}{t}^{\mathrm{4}} \:+\mathrm{10}{t}^{\mathrm{2}} \:+\mathrm{3}}{dt}\:{let}\:{solve}\:\mathrm{3}{t}^{\mathrm{4}} \:+\mathrm{10}{t}^{\mathrm{2}} \:+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{u}^{\mathrm{4}} \:+\mathrm{10}{u}\:+\mathrm{3}=\mathrm{0}\:\rightarrow\Delta^{'} =\mathrm{5}^{\mathrm{2}} −\mathrm{9}\:=\mathrm{16}\:\Rightarrow{u}_{\mathrm{1}} =\frac{−\mathrm{5}+\mathrm{4}}{\mathrm{3}}\:=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${u}_{\mathrm{2}} =\frac{−\mathrm{5}−\mathrm{4}}{\mathrm{3}}\:=−\mathrm{3}\:\Rightarrow\mathrm{3}{t}^{\mathrm{4}} \:+\mathrm{10}{t}^{\mathrm{2}} \:+\mathrm{3}\:=\mathrm{3}\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\right)\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)\:\Rightarrow \\ $$$${A}\:=\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left({arctant}\right)^{\mathrm{2}} }{\mathrm{3}\left({t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)}{dt}\:\Rightarrow \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\:{A}\:=\int_{−\infty} ^{\infty} \:\:\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left({arctant}\right)^{\mathrm{2}} }{\left({t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)}{dt}\:{letW}\left({z}\right)=\frac{\left(\mathrm{1}−{z}^{\mathrm{2}} \:\right)\left({arctanz}\right)^{\mathrm{2}} }{\left({z}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({z}^{\mathrm{2}} \:+\mathrm{3}\right)}\:\Rightarrow \\ $$$${W}\left({z}\right)\:=\frac{\left(\mathrm{1}−{z}^{\mathrm{2}} \right)\left({arctanz}\right)^{\mathrm{2}} }{\left({z}−\frac{{i}}{\:\sqrt{\mathrm{3}}}\right)\left({z}+\frac{{i}}{\:\sqrt{\mathrm{3}}}\right)\left({z}−{i}\sqrt{\mathrm{3}}\right)\left({z}+{i}\sqrt{\mathrm{3}}\right)}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left({W},\frac{{i}}{\:\sqrt{\mathrm{3}}}\right)+{Res}\left({W},{i}\sqrt{\mathrm{3}}\right)\right\} \\ $$$$…{be}\:{continued}… \\ $$$$ \\ $$$$ \\ $$

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