Question Number 218150 by Shrodinger last updated on 31/Mar/25 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}{x}}\right){dx} \\ $$ Answered by MrGaster last updated on 31/Mar/25 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}}…
Question Number 218148 by mnjuly1970 last updated on 30/Mar/25 $$ \\ $$$$\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{sin}\left(\sqrt{\:{x}\:}\right)}{\:\sqrt[{\mathrm{4}}]{\:{e}^{{x}} }}{dx}=? \\ $$$$ \\ $$ Answered by MrGaster last updated on…
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Question Number 217881 by OmoloyeMichael last updated on 23/Mar/25 Answered by vnm last updated on 23/Mar/25 $$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {x}\left(\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}\right){dx}={I} \\ $$$${f}\left({x}\right)=\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}…
Question Number 217813 by Wuji last updated on 21/Mar/25 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left[\left({xp}\left(\mathrm{2}+{x}\right)\right]^{−\mathrm{1}} {dx}\:\:\:\right. \\ $$$${p}\in\mathbb{R} \\ $$ Answered by mr W last updated on 22/Mar/25…
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Question Number 217755 by Tawa11 last updated on 20/Mar/25 $$\int\:\frac{\mathrm{cos}\left(\mathrm{sin}^{−\:\mathrm{1}} \mathrm{x}\right)\:+\:\mathrm{cos}^{−\:\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{1}\:+\:\sqrt{\mathrm{x}\:+\:\sqrt{\mathrm{x}}}\right)\right.\right.}\:\mathrm{dx} \\ $$ Commented by mr W last updated on 20/Mar/25 $${you}\:{can}\:{even}\:{make}\:{it}\:{more}\:{nice} \\ $$$${looking}…
Question Number 217761 by mnjuly1970 last updated on 20/Mar/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{prove}\:{that}\:: \\ $$$$ \\ $$$$ \\ $$$$\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\pi{x}\right){sin}\left(\mathrm{2}\pi{x}\right){sin}\left(\mathrm{3}\pi{x}\right)}{{x}^{\mathrm{3}} }\:=\:\pi^{\mathrm{3}} \:\:\:\:\:\:\:\:\: \\ $$$$ \\…
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Question Number 217683 by MrGaster last updated on 18/Mar/25 $$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }+\mathrm{6}{K}^{'} }{{K}^{\mathrm{2}} +\mathrm{36}{K}^{'\mathrm{2}} }\:}\frac{{dk}}{\:\sqrt{{k}}\left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} }=\sqrt{\pi}\left(\sqrt{\mathrm{2}}−\sqrt{\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}}\right) \\ $$ Answered by MrGaster last…