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Question Number 28159 by abdo imad last updated on 21/Jan/18

$$letgiveD=[0,\frac{\pi }{2}]\times [0,\frac{1}{2}]findthevalueof$$$$\int {\int }_D\frac{dxdy}{ycosx+1}.$$

Commented by abdo imad last updated on 26/Jan/18

$$letputI=\int {\int }_D\frac{dxdy}{ycosx+1}$$$$I={\int }_0^{\frac{\pi }{2}}\left({\int }_0^{\frac{1}{2}}\frac{dy}{ycox+1}\right)dx={\int }_0^{\frac{\pi }{2}}{\left[ln(ycosx+1)\right]}_{y=0}^{y=\frac{1}{2}}\frac{dx}{cosx}$$$$={\int }_0^{\frac{\pi }{2}}\frac{ln(1+\frac{1}{2}cosx)}{cosx}dxandbyfubini$$$$I={\int }_0^{\frac{1}{2}}\left({\int }_0^{\frac{\pi }{2}}\frac{dx}{ycosx+1}\right)dythech.tan\left(\frac{x}{2}\right)=tgive$$$${\int }_0^{\frac{\pi }{2}}\frac{dx}{ycosx+1}={\int }_0^1\frac{1}{1+y\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}$$$$={\int }_0^1\frac{2dt}{1+t^2+y(1-t^{2)}}={\int }_0^1\frac{2dt}{(1-y)t^2+1+y}$$$$=\frac{2}{1-y}{\int }_0^1\frac{dt}{t^2+\frac{1+y}{1-y}}(ch.t=\sqrt{\frac{1+y}{1-y}}u)$$$$=\frac{2}{1-y}{\int }_0^{\sqrt{\frac{1-y}{1+y}}}\frac{1}{\frac{1+y}{1-y}(1+u^2)}\sqrt{\frac{1+y}{1-y}}du$$$$=\frac{2}{1+y}.\frac{\sqrt{1+y}}{\sqrt{1-y}}artan\left(\sqrt{\frac{1-y}{1+y}}\right)=\frac{2}{\sqrt{1-y^2}}arctan\left(\sqrt{\frac{1-y}{1+y}}\right)$$$$=\frac{2}{sin\theta }arctan\left(tan\left(\frac{\theta }{2}\right)\right)(ch.y=cos\theta )$$$$=\frac{\theta }{sin\theta }=\frac{arcosy}{\sqrt{1-y^2}}$$$$\Rightarrow I={\int }_0^{\frac{1}{2}}\frac{arcosy}{\sqrt{1-y^2}}dybyparts$$$$I={[-arcosy.arcosy]}_0^{\frac{1}{2}}-{\int }_0^{\frac{1}{2}}-arcosy\frac{-dy}{\sqrt{1-y^2}}$$$$2I={\left(\frac{\pi }{2}\right)}^2-{\left(\frac{\pi }{3}\right)}^2=\frac{{\pi }^2}{4}-\frac{{\pi }^2}{9}=\frac{5{\pi }^2}{36}$$$$andI=\frac{5{\pi }^2}{72}.$$

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