Question Number 190672 by 073 last updated on 08/Apr/23
Answered by aba last updated on 08/Apr/23
$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{2}} }\right)\pi? \\ $$
Commented by 073 last updated on 08/Apr/23
$$\mathrm{solution}\:\mathrm{please}?? \\ $$
Answered by qaz last updated on 09/Apr/23
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}^{−\mathrm{1}} \left\{\int_{\mathrm{0}} ^{\infty} \frac{\mathscr{L}\left\{\mathrm{cos}\:{tx}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\right\}_{{t}=\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}^{−\mathrm{1}} \left\{\int_{\mathrm{0}} ^{\infty} \frac{{s}}{\left({s}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}\right\}_{{t}=\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}\mathscr{L}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{s}+\mathrm{1}}\right\}_{{t}=\mathrm{2}} =\frac{\pi}{\mathrm{4}}\left(\mathrm{1}+{e}^{−{t}} \right)_{{t}=\mathrm{2}} =\frac{\pi}{\mathrm{4}}\left(\mathrm{1}+{e}^{−\mathrm{2}} \right) \\ $$
Commented by aba last updated on 09/Apr/23
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