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Question Number 120316 by Bird last updated on 30/Oct/20

$$calculate{\int }_0^{\frac{\pi }{2}}\frac{xcosx}{cos\left(2x\right)}dx$$

Answered by mathmax by abdo last updated on 30/Oct/20

$$\mathrm{let}\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{x}}{\mathrm{cosx}+\mathrm{sinx}}\mathrm{dx}\mathrm{we}\mathrm{have}\mathrm{proved}\mathrm{that}\mathrm{I}=\frac{\pi }{4\sqrt{2}}\ln \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$$$$\mathrm{changement}\mathrm{x}=\mathrm{t}-\frac{\pi }{2}\mathrm{give}$$$$\mathrm{I}=-{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{t}-\frac{\pi }{2}}{\mathrm{sint}-\mathrm{cost}}\mathrm{dt}=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{\mathrm{sint}-\mathrm{cost}}-{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{t}}{\mathrm{sint}-\mathrm{cost}}$$$$={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{x}}{\mathrm{cosx}-\mathrm{sinx}}\mathrm{dx}+\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\mathrm{sinx}-\mathrm{cosx}}\Rightarrow$$$$2\mathrm{I}={\int }_0^{\frac{\pi }{2}}\mathrm{x}(\frac{1}{\mathrm{cosx}+\mathrm{sinx}}+\frac{1}{\mathrm{cosx}-\mathrm{sinx}})\mathrm{dx}+\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\mathrm{sinx}-\mathrm{cosx}}$$$$={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{x}\left(2\mathrm{cosx}\right)}{{\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}}\mathrm{dx}+\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\mathrm{sinx}-\mathrm{cosx}}$$$$\Rightarrow 2{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{xcosx}}{\cos \left(2\mathrm{x}\right)}\mathrm{dx}=2\mathrm{I}-\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\mathrm{sinx}-\mathrm{cosx}}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{xcosx}}{\cos \left(2\mathrm{x}\right)}\mathrm{dx}=\mathrm{I}-\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\mathrm{sinx}-\mathrm{cosx}}\mathrm{we}\mathrm{have}$$$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{\mathrm{sinx}-\mathrm{cosx}}{=}_{\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}}{\int }_0^1\frac{2\mathrm{dt}}{(1+{\mathrm{t}}^2)(\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}-\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2})}$$$$={\int }_0^1\frac{2\mathrm{dt}}{2\mathrm{t}-1+{\mathrm{t}}^2}=2{\int }_0^1\frac{\mathrm{dt}}{{\mathrm{t}}^2+2\mathrm{t}-1}=2{\int }_0^1\frac{\mathrm{dt}}{{(\mathrm{t}+1)}^2-2}$$$$=2{\int }_0^1\frac{\mathrm{dt}}{(\mathrm{t}+1-\sqrt{2})(\mathrm{t}+1+\sqrt{2})}=\frac{2}{2\sqrt{2}}{\int }_0^1(\frac{1}{\mathrm{t}+1-\sqrt{2}}-\frac{1}{\mathrm{t}+1+\sqrt{2}})$$$$=\frac{1}{\sqrt{2}}{[\ln \mid \frac{\mathrm{t}+1-\sqrt{2}}{\mathrm{t}+1+\sqrt{2}}\mid ]}_0^1=\frac{1}{\sqrt{2}}\ln \mid \frac{2-\sqrt{2}}{2+\sqrt{2}}\mid -\ln \mid \frac{1-\sqrt{2}}{1+\sqrt{2}}\mid$$$$=\frac{1}{\sqrt{2}}\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)-\ln \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{xcosx}}{\cos \left(2\mathrm{x}\right)}\mathrm{dx}=\frac{\pi }{4\sqrt{2}}\ln \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)-\frac{\pi }{4\sqrt{2}}\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)+\frac{\pi }{4}\ln \left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)$$$$=\frac{\pi }{4}(\frac{1}{\sqrt{2}}-1)\ln \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)-\frac{\pi }{4\sqrt{2}}\ln \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)$$

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