Question Number 64068 by mathmax by abdo last updated on 12/Jul/19
$${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{tsint}}{\mathrm{3}+{sin}^{\mathrm{2}} {t}}\:{dt}\: \\ $$
Commented by mathmax by abdo last updated on 17/Jul/19
![let A =∫_0 ^π ((tsint)/(3 +sin^2 t))dt changement t =π−x give A =−∫_0 ^π (((π−x)sin(π−x))/(3+sin^2 (π−x))) (−dx) =∫_0 ^π ((πsinx)/(3+sin^2 x))dx−∫_0 ^π ((xsinx)/(3+sin^2 x))dx ⇒2A =π ∫_0 ^π ((sinx)/(3+sin^2 x)) dx ⇒A =(π/2) ∫_0 ^π ((sinx )/(3+sin^2 x))dx changement tan((x/2)) =u give ∫_0 ^π ((sinx)/(3+sin^2 x))dx =∫_0 ^∞ (((2u)/(1+u^2 ))/(3 +((4u^2 )/((1+u^2 )^2 )))) ((2du)/(1+u^2 )) = ∫_0 ^∞ ((4u)/((1+u^2 )^2 {3 +((4u^2 )/((1+u^2 )^2 ))})) =∫_0 ^∞ ((4u)/(3(1+u^2 )^2 +4u^2 ))du =∫_0 ^∞ ((4u)/(3(u^4 +2u^2 +1) +4u^2 ))du =∫_0 ^∞ ((4u)/(3u^4 +6u^2 +3 +4u^2 ))du =∫_0 ^∞ ((4u)/(3u^4 +10u^2 +3))du 3u^4 +10 u^2 +3 =0 ⇒3x^2 +10x +3 =0 (x=u^2 ) Δ^′ =25−9 =16 ⇒x_1 =((−5 +4)/3) =−(1/3) x_2 =((−5−4)/3) =−3 ⇒3x^2 +10x +3 =3(x+(1/3))(x+3) ⇒ 3u^4 +10u^2 +3 =3(u^2 +(1/3))(u^2 +3) ⇒ ((4u)/(3u^4 +10u^2 +3)) =((4u)/3)(3/8){(1/(u^2 +(1/3)))−(1/(u^2 +3))} =(u/2){ (1/(u^2 +(1/3)))−(1/(u^2 +3))} ⇒ ∫_0 ^∞ ((4u)/(3u^4 +10u^2 +3))du =(1/2){ ∫_0 ^∞ ((udu)/(u^2 +(1/3))) −∫_0 ^∞ ((u du)/(u^2 +3))} =(1/4)[ln∣((u^2 +(1/3))/(u^2 +3))∣]_0 ^(+∞) =(1/4)(−ln((1/9)))=((ln9)/4) =((2ln3)/4) =((ln3)/2) ⇒ A =(π/2) ((ln3)/2) ⇒ A =((πln3)/4)](https://www.tinkutara.com/question/Q64402.png)
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{tsint}}{\mathrm{3}\:+{sin}^{\mathrm{2}} {t}}{dt}\:{changement}\:{t}\:=\pi−{x}\:{give} \\ $$$${A}\:=−\int_{\mathrm{0}} ^{\pi} \:\:\frac{\left(\pi−{x}\right){sin}\left(\pi−{x}\right)}{\mathrm{3}+{sin}^{\mathrm{2}} \left(\pi−{x}\right)}\:\left(−{dx}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\pi{sinx}}{\mathrm{3}+{sin}^{\mathrm{2}} {x}}{dx}−\int_{\mathrm{0}} ^{\pi} \:\frac{{xsinx}}{\mathrm{3}+{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\Rightarrow\mathrm{2}{A}\:=\pi\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sinx}}{\mathrm{3}+{sin}^{\mathrm{2}} {x}}\:{dx}\:\Rightarrow{A}\:=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sinx}\:}{\mathrm{3}+{sin}^{\mathrm{2}} {x}}{dx} \\ $$$${changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:={u}\:{give}\: \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sinx}}{\mathrm{3}+{sin}^{\mathrm{2}} {x}}{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{3}\:+\frac{\mathrm{4}{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{4}{u}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} \left\{\mathrm{3}\:+\frac{\mathrm{4}{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{4}{u}}{\mathrm{3}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{4}{u}^{\mathrm{2}} }{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{4}{u}}{\mathrm{3}\left({u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}\right)\:+\mathrm{4}{u}^{\mathrm{2}} }{du}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{4}{u}}{\mathrm{3}{u}^{\mathrm{4}} \:+\mathrm{6}{u}^{\mathrm{2}} \:+\mathrm{3}\:+\mathrm{4}{u}^{\mathrm{2}} }{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{4}{u}}{\mathrm{3}{u}^{\mathrm{4}} \:+\mathrm{10}{u}^{\mathrm{2}} \:+\mathrm{3}}{du} \\ $$$$\mathrm{3}{u}^{\mathrm{4}} \:+\mathrm{10}\:{u}^{\mathrm{2}} \:+\mathrm{3}\:=\mathrm{0}\:\Rightarrow\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{10}{x}\:+\mathrm{3}\:=\mathrm{0}\:\:\:\left({x}={u}^{\mathrm{2}} \right) \\ $$$$\Delta^{'} \:=\mathrm{25}−\mathrm{9}\:=\mathrm{16}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{5}\:+\mathrm{4}}{\mathrm{3}}\:=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}_{\mathrm{2}} =\frac{−\mathrm{5}−\mathrm{4}}{\mathrm{3}}\:=−\mathrm{3}\:\Rightarrow\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{10}{x}\:+\mathrm{3}\:=\mathrm{3}\left({x}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({x}+\mathrm{3}\right)\:\Rightarrow \\ $$$$\mathrm{3}{u}^{\mathrm{4}} \:+\mathrm{10}{u}^{\mathrm{2}} \:+\mathrm{3}\:=\mathrm{3}\left({u}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({u}^{\mathrm{2}} \:+\mathrm{3}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{4}{u}}{\mathrm{3}{u}^{\mathrm{4}} \:+\mathrm{10}{u}^{\mathrm{2}} \:+\mathrm{3}}\:=\frac{\mathrm{4}{u}}{\mathrm{3}}\frac{\mathrm{3}}{\mathrm{8}}\left\{\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{3}}\right\}\:\:\:\:=\frac{{u}}{\mathrm{2}}\left\{\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{3}}\right\}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{4}{u}}{\mathrm{3}{u}^{\mathrm{4}} \:+\mathrm{10}{u}^{\mathrm{2}} \:+\mathrm{3}}{du}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{udu}}{{u}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}\:−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{u}\:{du}}{{u}^{\mathrm{2}} \:+\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[{ln}\mid\frac{{u}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{3}}}{{u}^{\mathrm{2}} \:+\mathrm{3}}\mid\right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{\mathrm{4}}\left(−{ln}\left(\frac{\mathrm{1}}{\mathrm{9}}\right)\right)=\frac{{ln}\mathrm{9}}{\mathrm{4}}\:=\frac{\mathrm{2}{ln}\mathrm{3}}{\mathrm{4}}\:=\frac{{ln}\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$${A}\:=\frac{\pi}{\mathrm{2}}\:\frac{{ln}\mathrm{3}}{\mathrm{2}}\:\:\Rightarrow\:{A}\:=\frac{\pi{ln}\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$ \\ $$