Question Number 28159 by abdo imad last updated on 21/Jan/18
![let give D=[0,(π/2)]×[0,(1/2)] find the value of ∫∫_D ((dxdy)/(ycosx +1)) .](https://www.tinkutara.com/question/Q28159.png)
$${let}\:{give}\:{D}=\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]×\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right]\:\:{find}\:{the}\:{value}\:{of} \\ $$$$\int\int_{{D}} \:\:\:\frac{{dxdy}}{{ycosx}\:+\mathrm{1}}\:\:. \\ $$
Commented by abdo imad last updated on 26/Jan/18
![let put I=∫∫_D ((dxdy)/(ycosx +1)) I= ∫_0 ^(π/2) ( ∫_0 ^(1/2) (dy/(ycox+1)))dx = ∫_0 ^(π/2) [ ln(ycosx+1)]_(y=0) ^(y=(1/2)) (dx/(cosx)) = ∫_0 ^(π/2) ((ln(1+(1/2)cosx))/(cosx))dx and by fubini I= ∫_0 ^(1/2) ( ∫_0 ^(π/2) (dx/(ycosx +1)))dy the ch. tan((x/2))=t give ∫_0 ^(π/2) (dx/(ycosx+1))= ∫_0 ^1 (1/(1+y((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^1 ((2dt)/(1+t^2 +y(1−t^(2)) ))= ∫_0 ^1 ((2dt)/((1−y)t^2 +1+y)) = (2/(1−y)) ∫_0 ^1 (dt/(t^2 +((1+y)/(1−y)))) ( ch.t=(√((1+y)/(1−y)))u) = (2/(1−y)) ∫_0 ^(√((1−y)/(1+y))) (1/(((1+y)/(1−y))(1+u^2 )))(√( ((1+y)/(1−y))))du = (2/(1+y)) .((√(1+y))/( (√(1−y)))) artan((√((1−y)/(1+y))))= (2/( (√(1−y^2 )))) arctan((√((1−y)/(1+y)))) = (2/(sinθ)) arctan(tan((θ/2))) (ch.y=cosθ) =(θ/(sinθ)) =((arcosy)/( (√(1−y^2 )))) ⇒ I= ∫_0 ^(1/2) ((arcosy)/( (√(1−y^2 ))))dy by parts I= [−arcosy.arcosy]_0 ^(1/2) − ∫_0 ^(1/2) −arcosy ((−dy)/( (√(1−y^2 )))) 2I= ((π/2))^2 −((π/3))^2 = (π^2 /4)− (π^2 /9)= ((5π^2 )/(36)) and I= ((5π^2 )/(72)) .](https://www.tinkutara.com/question/Q28470.png)
$${let}\:{put}\:\:{I}=\int\int_{{D}} \:\frac{{dxdy}}{{ycosx}\:+\mathrm{1}} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\:\frac{{dy}}{{ycox}+\mathrm{1}}\right){dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\:{ln}\left({ycosx}+\mathrm{1}\right)\right]_{{y}=\mathrm{0}} ^{{y}=\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dx}}{{cosx}} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{cosx}\right)}{{cosx}}{dx}\:\:{and}\:{by}\:{fubini} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{{ycosx}\:+\mathrm{1}}\right){dy}\:{the}\:{ch}.\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{{ycosx}+\mathrm{1}}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+{y}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+{y}\left(\mathrm{1}−{t}^{\left.\mathrm{2}\right)} \right.}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}−{y}\right){t}^{\mathrm{2}} +\mathrm{1}+{y}} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{1}−{y}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}}}\:\:\:\:\:\:\left(\:\:\:\:\:\:{ch}.{t}=\sqrt{\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}}}{u}\right) \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{1}−{y}}\:\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}}} \:\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\sqrt{\:\:\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}}}{du} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{1}+{y}}\:.\frac{\sqrt{\mathrm{1}+{y}}}{\:\sqrt{\mathrm{1}−{y}}}\:{artan}\left(\sqrt{\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}}\right)=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}}\right) \\ $$$$=\:\frac{\mathrm{2}}{{sin}\theta}\:{arctan}\left({tan}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:\:\:\left({ch}.{y}={cos}\theta\right) \\ $$$$=\frac{\theta}{{sin}\theta}\:=\frac{{arcosy}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \\ $$$$\Rightarrow\:{I}=\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\frac{{arcosy}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{dy}\:\:\:{by}\:{parts} \\ $$$${I}=\:\:\left[−{arcosy}.{arcosy}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:−\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} −{arcosy}\:\frac{−{dy}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \\ $$$$\mathrm{2}{I}=\:\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \:−\left(\frac{\pi}{\mathrm{3}}\right)^{\mathrm{2}} =\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\:\frac{\pi^{\mathrm{2}} }{\mathrm{9}}=\:\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$${and}\:\:{I}=\:\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{72}}\:. \\ $$