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Question-219425

Question Number 219425 by Nicholas666 last updated on 24/Apr/25 Commented by Nicholas666 last updated on 24/Apr/25 $$\:\:\:\:{ABC}\:{reguler}\:{pentagon}\: \\ $$$$\:\:\:{P},\:{Q},\:{T}\:\:\:\:\:{toricelli}'{s}\:{point}\:{of}\:{AED},{BCD},{ADE}\:\:\:\:\: \\ $$$$\:\:\:{find}\:{angle}\:{PTQ}? \\ $$ Answered by…

0-sin-2-u-u-2-du-I-I-t-0-sin-2-u-u-2-e-ut-du-0-1-u-sin-2-u-u-e-ut-du-t-L-u-sin-2-p-p-du-t-0-sin-

Question Number 219388 by SdC355 last updated on 23/Apr/25 $$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }\:\mathrm{d}{u}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }{e}^{−{ut}} \mathrm{d}{u}=\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{1}}{{u}}\centerdot\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}}{e}^{−{ut}} \mathrm{d}{u}=…

Question-219384

Question Number 219384 by mnjuly1970 last updated on 23/Apr/25 Answered by SdC355 last updated on 24/Apr/25 $$\int\:\:\:\frac{\mathrm{d}{x}}{\:\sqrt{{x}}}\:\mathrm{cos}^{\mathrm{3}} \left({x}\right)\mathrm{sin}\left({x}\right)={I} \\ $$$$\int\:\:\:\frac{\mathrm{d}{x}}{\:{x}}\:\sqrt{{x}}\mathrm{cos}^{\mathrm{3}} \left({x}\right)\mathrm{sin}\left({x}\right)=\int\:\:\:\frac{\mathrm{d}{x}}{{x}}\:\sqrt{{x}}\mathrm{cos}^{\mathrm{2}} \left({x}\right)\mathrm{cos}\left({x}\right)\mathrm{sin}\left({x}\right) \\ $$$$\int\:\:\frac{\mathrm{d}{x}}{\:\mathrm{2}{x}}\:\sqrt{{x}}\mathrm{cos}^{\mathrm{2}} \left({x}\right)\mathrm{sin}\left(\mathrm{2}{x}\right)=\int\:\:\mathrm{d}{x}\:\frac{{e}^{−{xt}}…

f-s-1-2pi-e-it-s-dt-0-e-it-s-e-sp-dtds-

Question Number 219323 by SdC355 last updated on 23/Apr/25 $${f}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int\:\:{e}^{−\boldsymbol{{i}}{t}\left({s}−\alpha\right)} \:\mathrm{d}{t}\: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \int_{−\infty} ^{\:+\infty} \:\:{e}^{−\boldsymbol{{i}}{t}\left({s}−\alpha\right)} {e}^{−{sp}} \mathrm{d}{t}\mathrm{d}{s}=? \\ $$ Terms of Service Privacy…