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Question Number 205279 by gopikrishnan last updated on 14/Mar/24

$${\underset{-\pi /2}{\int }}^{\pi /2}\frac{8\sqrt{2}cosx}{(1+\stackrel{sinx}{e})(1+si\stackrel{4}{nx})}dx=a\pi +blog(3+2\sqrt{2})thenfinda+b$$

Answered by Berbere last updated on 14/Mar/24

$$x\to -x$$$$\mathrm{\Omega }={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{8\sqrt{2}cos\left(x\right)}{(1+e^{sin\left(x\right)})(1+{sin}^4\left(x\right))}dx={\int }_{\frac{\pi }{2}}^{-\frac{\pi }{2}}-\frac{8\sqrt{2}cos(-x)}{(1+e^{-sin\left(x\right)})(1+{sin}^4\left(x\right))}dx$$$$=\mathrm{\Omega }$$$$2\mathrm{\Omega }={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{8\sqrt{2}cos\left(x\right)}{1+{sin}^4\left(x\right)};sin\left(x\right)=y$$$$2\mathrm{\Omega }={\int }_{-1}^1\frac{8\sqrt{2}dy}{1+y^4}\Rightarrow \mathrm{\Omega }={\int }_0^1\frac{8\sqrt{2}dy}{(y^2+1+y\sqrt{2})(y^2+1-y\sqrt{2})}$$$$\frac{8\sqrt{2}}{(y^4+1)}=\frac{ay+b}{y^2+1+y\sqrt{2}}+\frac{cy+d}{y^2+1-y\sqrt{2}}$$$$a+c=0,(-a\sqrt{2}+b+c\sqrt{2}+d)=0$$$$b+d=8\sqrt{2};(a+c+d\sqrt{2}-b\sqrt{2})=0$$$$a-c=8$$$$d-b=0$$$$d=b=4\sqrt{2};a=4,c=-4$$$$\frac{4y+4\sqrt{2}}{y^2+y\sqrt{2}+1}+\frac{-4y+4\sqrt{2}}{y^2+1-y\sqrt{2}}$$$$\frac{2(2y+\sqrt{2})}{y^2+y\sqrt{2}+1}+\frac{2\sqrt{2}}{{(y+\frac{1}{\sqrt{2}})}^2+\frac{1}{2}}+\frac{-2(2y-\sqrt{2})}{y^2+1-y\sqrt{2}}+\frac{2\sqrt{2}}{{(y-\frac{1}{2})}^2+\frac{1}{2}}$$$$\int \frac{8\sqrt{2}}{y^4+1}dy=2ln\left(\frac{y^2+y\sqrt{2}+1}{y^2-y\sqrt{2}+1}\right)+{4\tan }^{-1}(y\sqrt{2}+1)+{4\tan }^{-1}(y\sqrt{2}-1)+c$$$$\mathrm{\Omega }=2ln\left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)+4({\tan }^{-1}(\sqrt{2}+1)+{\tan }^{-1}(\sqrt{2}-1)-{\tan }^{-1}\left(1\right)-{\tan }^{-1}(-1))$$$$=2ln\left(\frac{6+4\sqrt{2}}{2}\right)+4({\tan }^{-1}(\sqrt{2}+1)+{\tan }^{-1}\left(\frac{1}{\sqrt{2}+1}\right))$$$$2ln(3+2\sqrt{2})+4.\frac{\pi }{2}=2\pi +2ln(3+2\sqrt{2)}$$

Commented by gopikrishnan last updated on 16/Mar/24

$$thanku$$

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