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Question-218812

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Question Number 218812 by Spillover last updated on 15/Apr/25 Answered by A5T last updated on 16/Apr/25 $$\sqrt{\left(\mathrm{R}−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{x}^{\mathrm{2}} −\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\mathrm{R}−\mathrm{2x}\right)\mathrm{R}=\mathrm{h}\left(\mathrm{2x}−\mathrm{h}\right) \\ $$$$\Rightarrow\mathrm{R}^{\mathrm{2}}…

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0-x-2-cos-x-cosh-2x-cos-x-2x-2-e-4x-2e-2x-cos-x-1-dx-lemma-k-1-cos-kx-p-k-p-cos-x-1-p-2-2p-cos-x-1-p-gt-1-

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Question Number 218748 by MrGaster last updated on 15/Apr/25 $$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{cos}\:{x}}{\mathrm{cosh}\:\mathrm{2}{x}−\mathrm{cos}\:{x}}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{e}^{\mathrm{4}{x}} −\mathrm{2}{e}^{\mathrm{2}{x}} \mathrm{cos}\:{x}+\mathrm{1}}{dx},\mathrm{lemma}:\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{cos}\:{kx}}{{p}^{{k}} }=\frac{\mathrm{p}\:\mathrm{cos}\:{x}−\mathrm{1}}{{p}^{\mathrm{2}} −\mathrm{2}{p}\:\mathrm{cos}\:{x}+\mathrm{1}},{p}>\mathrm{1} \\ $$ Answered by MrGaster…

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