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Question Number 30180 by abdo imad last updated on 17/Feb/18
find ∫_0 ^(π/2) ((x sinx cosx)/(tan^2 x +cotan^2 x))dx .(use the ch.x=(π/2) −t).
$${find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{x}\:{sinx}\:{cosx}}{{tan}^{\mathrm{2}} {x}\:+{cotan}^{\mathrm{2}} {x}}{dx}\:.\left({use}\:{the}\:{ch}.{x}=\frac{\pi}{\mathrm{2}}\:−{t}\right). \\ $$
Commented by abdo imad last updated on 24/Feb/18
ch.x=(π/2) −t give I= ∫_0 ^(π/2) ((((π/2)−t)cost sint)/(cotan^2 t +tan^2 t))dt = (π/2)∫_0 ^(π/2) (dt/(tan^2 t +cotan^2 t)) −∫_0 ^(π/2) ((sint?cost)/(tan^2 t +cotan^2 t))dt ⇒ 2I= (π/2) ∫_0 ^(π/2) (dt/(tan^2 t +cotan^2 t)) dt the ch. tant= u give ∫_0 ^(π/2) (dt/(tan^2 t +cotan^2 t))= ∫_0 ^∞ (1/(u^(2 ) +(1/u^2 ))) (du/(1+u^2 )) = ∫_0 ^∞ (u^2 /((1+u^2 )(1+u^4 )))du= ∫_0 ^∞ ((1+u^2 −1)/((1+u^2 )(1+u^4 )))du = ∫_0 ^∞ (du/(1+u^4 )) −∫_0 ^∞ (du/((1+u^2 )(1+u^4 ))) the ch. u^4 =x give ∫_0 ^∞ (du/(1+u^4 )) = ∫_0 ^∞ (((1/4)x^((1/4)−1) )/(1+x))dx= (1/4) (π/(sin((π/4)))) = (π/4) (1/(1/( (√2))))= ((π(√2))/4) .let decompose F(x)= (1/((1+u^2 )(1+u^4 ))) F(x)= (1/((1+u^2 )( (u^2 +1)^2 −2u^2 ))) = (1/((1+u^2 )(u^2 +1 −(√2)u)( u^2 +1 +(√2) u))) = ((au+b)/(1+u^2 )) +((cu +d)/(u^2 −(√2)u +1)) + ((eu +f)/(u^2 + (√2)u +1)) ....be continued...
$${ch}.{x}=\frac{\pi}{\mathrm{2}}\:−{t}\:{give}\:\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\frac{\pi}{\mathrm{2}}−{t}\right){cost}\:{sint}}{{cotan}^{\mathrm{2}} {t}\:+{tan}^{\mathrm{2}} {t}}{dt} \\ $$$$=\:\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{{tan}^{\mathrm{2}} {t}\:+{cotan}^{\mathrm{2}} {t}}\:\:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{sint}?{cost}}{{tan}^{\mathrm{2}} {t}\:+{cotan}^{\mathrm{2}} {t}}{dt}\:\:\Rightarrow \\ $$$$\mathrm{2}{I}=\:\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{{tan}^{\mathrm{2}} {t}\:+{cotan}^{\mathrm{2}} {t}}\:{dt}\:\:{the}\:{ch}.\:{tant}=\:{u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{{tan}^{\mathrm{2}} {t}\:+{cotan}^{\mathrm{2}} {t}}=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}\:} \:+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}{du}=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}+{u}^{\mathrm{2}} \:−\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}{du} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{4}} }\:\:−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}\:{the}\:{ch}.\:{u}^{\mathrm{4}} ={x}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{4}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\frac{\mathrm{1}}{\mathrm{4}}{x}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{x}}{dx}=\:\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\:\frac{\pi}{\mathrm{4}}\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:.{let}\:{decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{4}} \right)} \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\:\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}{u}^{\mathrm{2}} \right)} \\ $$$$=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\:−\sqrt{\mathrm{2}}{u}\right)\left(\:{u}^{\mathrm{2}} \:+\mathrm{1}\:+\sqrt{\mathrm{2}}\:{u}\right)} \\ $$$$=\:\frac{{au}+{b}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:+\:\frac{{eu}\:+{f}}{{u}^{\mathrm{2}} \:+\:\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:….{be}\:{continued}… \\ $$

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