Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 86613 by Ar Brandon last updated on 29/Mar/20

$${\int }_0^{\frac{1}{2}}{\int }_0^{\frac{\pi }{2}}\frac{1}{ycos\left(x\right)+1}dxdy$$$$$$

Commented by abdomathmax last updated on 29/Mar/20

$${\int }_0^{\frac{\pi }{2}}\frac{dx}{ycosx+1}{=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^1\frac{2dt}{(1+t^2)(y\frac{1-t^2}{1+t^2}+1)}$$$$=2{\int }_0^1\frac{dt}{y-{yt}^2+1+t^2}=2{\int }_0^1\frac{dy}{(1-y)t^2+1+y}$$$$=\frac{2}{(1-y)}{\int }_0^1\frac{dy}{t^2+\frac{1+y}{1-y}}$$$${=}_{t=\sqrt{\frac{1+y}{1-y}}u}\frac{2}{(1-y)}\times \frac{1-y}{1+y}{\int }_0^{\sqrt{\frac{1-y}{1+y}}}\frac{1}{1+u^2}\times \sqrt{\frac{1+y}{1-y}}du$$$$=\frac{2}{\sqrt{1-y^2}}arctan\left(\sqrt{\frac{1-y}{1+y}}\right)\Rightarrow$$$${\int }_0^{\frac{1}{2}}{\int }_0^{\frac{\pi }{2}}\frac{dxdy}{ycosx+1}=2{\int }_0^{\frac{1}{2}}\frac{1}{\sqrt{1-y^2}}arctan\left(\sqrt{\frac{1-y}{1+y}}\right)dy$$$$...becontinued...$$

Commented by mind is power last updated on 30/Mar/20

$$y=cos\left(2\theta \right)$$$$aarctan\left(\sqrt{\frac{1-y}{1+y}}\right)=arctan\left(tan\left(\theta \right)\right)=\theta$$

Commented by mathmax by abdo last updated on 30/Mar/20

$$changementy=cos\theta give$$$${\int }_0^{\frac{1}{2}}\frac{1}{\sqrt{1-y^2}}arctan\left(\sqrt{\frac{1-y}{1+y}}\right)dy={\int }_{\frac{\pi }{2}}^{\frac{\pi }{3}}\frac{1}{sin\theta }arctan\left(\sqrt{\frac{2{sin}^2\left(\frac{\theta }{2}\right)}{2{cos}^2\left(\frac{\theta }{2}\right)}}\right)(-sin\theta )d\theta$$$$={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}arctan\left(tan\left(\frac{\theta }{2}\right)\right)d\theta =\frac{1}{2}{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\theta d\theta =\frac{1}{2}{\left[\frac{{\theta }^2}{2}\right]}_{\frac{\pi }{3}}^{\frac{\pi }{2}}$$$$=\frac{1}{4}(\frac{{\pi }^2}{4}-\frac{{\pi }^2}{9})=\frac{1}{4}\left(\frac{9{\pi }^2-4{\pi }^2}{36}\right)=\frac{5{\pi }^2}{144}$$

Commented by mathmax by abdo last updated on 30/Mar/20

$$thankyousirmind..$$

Commented by Ar Brandon last updated on 30/Mar/20

$$greatjob$$

Commented by abdomathmax last updated on 30/Mar/20

$$youarewelcomesir.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com