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Category: Integration

prove-that-x-2-e-x-2-cos-x-2-sin-x-2-dx-pi-sin-3-tan-1-2-2-4-125-1-4-

Question Number 129764 by Eric002 last updated on 18/Jan/21 $${prove}\:{that} \\ $$$$\int_{−\infty} ^{+\infty} {x}^{\mathrm{2}} \:{e}^{−{x}^{\mathrm{2}} } \:{cos}\left({x}^{\mathrm{2}} \right){sin}\left({x}^{\mathrm{2}} \right)\:{dx} \\ $$$$=\frac{\sqrt{\pi}{sin}\left[\frac{\sqrt{\mathrm{3}}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)}{\mathrm{2}}\right]}{\mathrm{4}\:\sqrt[{\mathrm{4}}]{\mathrm{125}}} \\ $$ Answered…

1-x-dx-

Question Number 64213 by Tony Lin last updated on 15/Jul/19 $$\int\frac{\mathrm{1}}{{x}!}{dx}=? \\ $$ Commented by MJS last updated on 15/Jul/19 $$\frac{\mathrm{1}}{{x}!}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{N}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{continuous} \\ $$$$\mathrm{for}\:{x}\in\mathbb{R}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{integrable} \\ $$…

N-3-2cos-x-2-3cos-x-dx-

Question Number 129746 by bramlexs22 last updated on 18/Jan/21 $$\:\mathrm{N}\:=\:\int\:\frac{\mathrm{3}+\mathrm{2cos}\:\mathrm{x}}{\mathrm{2}+\mathrm{3cos}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$ Answered by liberty last updated on 18/Jan/21 $$\:\mathrm{N}=\:\int\:\frac{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}+\mathrm{3cos}\:\mathrm{x}\right)+\frac{\mathrm{8}}{\mathrm{3}}}{\mathrm{2}+\mathrm{3cos}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$$$\:\mathrm{N}=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}\:+\:\frac{\mathrm{8}}{\mathrm{3}}\int\:\frac{\mathrm{dx}}{\mathrm{2}+\mathrm{3}\left(\mathrm{2cos}\:^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}−\mathrm{1}\right)} \\ $$$$\:\mathrm{N}=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}+\frac{\mathrm{8}}{\mathrm{3}}\int\:\frac{\mathrm{dx}}{\mathrm{6cos}\:^{\mathrm{2}}…

Question-64200

Question Number 64200 by aliesam last updated on 15/Jul/19 Commented by mathmax by abdo last updated on 15/Jul/19 $${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}−\mathrm{1}}{{lnx}}{dx}\:{changement}\:{lnx}=−{t}\:{give}\:{x}={e}^{−{t}} \\ $$$${I}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}}…

Question-64176

Question Number 64176 by aliesam last updated on 15/Jul/19 Answered by MJS last updated on 15/Jul/19 $$\mathrm{the}\:\mathrm{minimum}\:\mathrm{of}\:{f}\left({x}\right)=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\mathrm{is}\:{f}\left(\mathrm{0}\right)=\mathrm{1}\:\Rightarrow\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{between}\:\mathrm{the}\:{x}−\mathrm{axis}\:\mathrm{and}\:{f}\left({x}\right)\:\mathrm{in}\:\left[−\mathrm{1};\:\mathrm{1}\right] \\ $$$$\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2}\:\Rightarrow\:\mathrm{2}<\underset{−\mathrm{1}} {\int}^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}…