Question Number 134636 by bobhans last updated on 06/Mar/21
$$\mathcal{INTEGRAL} \\ $$$$\left(\mathrm{1}\right)\int_{\mathrm{0}} ^{\:\mathrm{ln}\:\mathrm{2}} \:{x}^{−\mathrm{2}} .{e}^{−\frac{\mathrm{1}}{{x}}} \:{dx}\:=? \\ $$$$\left(\mathrm{2}\right)\:\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}\:\mathrm{dx}\:=? \\ $$
Answered by benjo_mathlover last updated on 06/Mar/21
![(1) ∫_0 ^(ln 2) (e^(−(1/x)) /x^2 ) dx = ∫_0 ^(ln 2) e^(−(1/x)) d(−(1/x)) = lim_(p→0) [ e^(−(1/x)) ]_p ^(ln 2) = e^(−(1/(ln 2))) −e^(−∞) = e^(−(1/(ln 2))) = (1/( (( e))^(1/(ln 2 )) ))](https://www.tinkutara.com/question/Q134642.png)
$$\left(\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{ln}\:\mathrm{2}} \:\frac{\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{ln}\:\mathrm{2}} \:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \:\mathrm{d}\left(−\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$=\:\underset{\mathrm{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \:\right]_{\mathrm{p}} ^{\mathrm{ln}\:\mathrm{2}} =\:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}} −\mathrm{e}^{−\infty} \\ $$$$=\:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{ln}\:\mathrm{2}\:}]{\:\mathrm{e}}}\: \\ $$
Answered by rs4089 last updated on 06/Mar/21
$$\int\frac{{sinx}+{cosx}}{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}}{dx} \\ $$$$\mathrm{2}\int\frac{{sinx}+{cosx}}{\mathrm{2}−{sin}^{\mathrm{2}} \mathrm{2}{x}}{dx} \\ $$$${let}\:{sinx}−{cosx}={t}\:\Rightarrow\left({sinx}+{cosx}\right){dx}={dt} \\ $$$${and}\:\:{sin}\mathrm{2}{x}=\mathrm{1}−{t}^{\mathrm{2}} \\ $$$$\mathrm{2}\int\frac{{dt}}{\mathrm{2}−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{2}\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\mathrm{2}{tan}^{−\mathrm{1}} {t}\:+{c} \\ $$$$=\mathrm{2}{tan}^{−\mathrm{1}} \left({sinx}−{cosx}\right)+{C} \\ $$
Answered by john_santu last updated on 06/Mar/21
$$\left(\mathrm{2}\right)\:\int\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}\:{dx}\: \\ $$$$=\:\int\:\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}+\pi/\mathrm{4}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\:{dx} \\ $$$$=\:\mathrm{2}\sqrt{\mathrm{2}}\:\int\:\frac{\mathrm{sin}\:\left({x}+\pi/\mathrm{4}\right)}{\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\:{dx} \\ $$$${substitute}\:{h}\:=\:{x}+\pi/\mathrm{4}\:{and}\:\mathrm{2}{x}\:=\:\mathrm{2}{h}−\pi/\mathrm{2} \\ $$$${I}=\mathrm{2}\sqrt{\mathrm{2}}\:\int\:\frac{\mathrm{sin}\:{h}}{\mathrm{2}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{h}}\:{dh} \\ $$$${letting}\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:{h}\:=\:{u}\: \\ $$$${I}=−\mathrm{2}\int\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\:−\mathrm{2arctan}\:\left({u}\right)+\:{c} \\ $$$$=\:−\mathrm{2arctan}\:\left(\sqrt{\mathrm{2}}\:\mathrm{cos}\:\left({x}+\pi/\mathrm{4}\right)\right)+\:{c}\: \\ $$
Answered by mathmax by abdo last updated on 27/Mar/21
![1) we have (d/dx)(e^(−(1/x)) ) =(1/x^2 )e^(−(1/x)) ⇒ ∫_0 ^(ln2) x^(−2) e^(−(1/x)) ex =[e^(−(1/x)) ]_0 ^(ln2) =e^(−(1/(ln2))) −0 =e^(−(1/(ln2)))](https://www.tinkutara.com/question/Q136856.png)
$$\left.\mathrm{1}\right)\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{ln2}} \:\mathrm{x}^{−\mathrm{2}} \:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \:\mathrm{ex}\:=\left[\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \right]_{\mathrm{0}} ^{\mathrm{ln2}} \:=\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{ln2}}} −\mathrm{0}\:=\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{ln2}}} \\ $$