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Question Number 134636 by bobhans last updated on 06/Mar/21

INTEGRAL (1)∫_0 ^( ln 2) x^(−2) .e^(−(1/x)) dx =? (2) ∫ ((sin x+cos x)/(sin^4 x+cos^4 x)) dx =?

$$\mathcal{INTEGRAL} \\ $$$$\left(\mathrm{1}\right)\int_{\mathrm{0}} ^{\:\mathrm{ln}\:\mathrm{2}} \:{x}^{−\mathrm{2}} .{e}^{−\frac{\mathrm{1}}{{x}}} \:{dx}\:=? \\ $$$$\left(\mathrm{2}\right)\:\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}\:\mathrm{dx}\:=? \\ $$

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 )) ))

$$\left(\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{ln}\:\mathrm{2}} \:\frac{\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{ln}\:\mathrm{2}} \:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \:\mathrm{d}\left(−\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$=\:\underset{\mathrm{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \:\right]_{\mathrm{p}} ^{\mathrm{ln}\:\mathrm{2}} =\:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}} −\mathrm{e}^{−\infty} \\ $$$$=\:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{ln}\:\mathrm{2}\:}]{\:\mathrm{e}}}\: \\ $$

Answered by rs4089 last updated on 06/Mar/21

∫((sinx+cosx)/(1−2sin^2 xcos^2 x))dx 2∫((sinx+cosx)/(2−sin^2 2x))dx let sinx−cosx=t ⇒(sinx+cosx)dx=dt and sin2x=1−t^2 2∫(dt/(2−(1−t^2 ))) 2∫(dt/(1+t^2 )) =2tan^(−1) t +c =2tan^(−1) (sinx−cosx)+C

$$\int\frac{{sinx}+{cosx}}{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}}{dx} \\ $$$$\mathrm{2}\int\frac{{sinx}+{cosx}}{\mathrm{2}−{sin}^{\mathrm{2}} \mathrm{2}{x}}{dx} \\ $$$${let}\:{sinx}−{cosx}={t}\:\Rightarrow\left({sinx}+{cosx}\right){dx}={dt} \\ $$$${and}\:\:{sin}\mathrm{2}{x}=\mathrm{1}−{t}^{\mathrm{2}} \\ $$$$\mathrm{2}\int\frac{{dt}}{\mathrm{2}−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{2}\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\mathrm{2}{tan}^{−\mathrm{1}} {t}\:+{c} \\ $$$$=\mathrm{2}{tan}^{−\mathrm{1}} \left({sinx}−{cosx}\right)+{C} \\ $$

Answered by john_santu last updated on 06/Mar/21

(2) ∫ ((sin x+cos x)/(sin^4 x+cos^4 x)) dx = ∫ (((√2) sin (x+π/4))/(1−(1/2)sin^2 2x)) dx = 2(√2) ∫ ((sin (x+π/4))/(2−sin^2 2x)) dx substitute h = x+π/4 and 2x = 2h−π/2 I=2(√2) ∫ ((sin h)/(2−cos^2 2h)) dh letting (√2) cos h = u I=−2∫ (du/(1+u^2 )) = −2arctan (u)+ c = −2arctan ((√2) cos (x+π/4))+ c

$$\left(\mathrm{2}\right)\:\int\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}\:{dx}\: \\ $$$$=\:\int\:\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}+\pi/\mathrm{4}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\:{dx} \\ $$$$=\:\mathrm{2}\sqrt{\mathrm{2}}\:\int\:\frac{\mathrm{sin}\:\left({x}+\pi/\mathrm{4}\right)}{\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\:{dx} \\ $$$${substitute}\:{h}\:=\:{x}+\pi/\mathrm{4}\:{and}\:\mathrm{2}{x}\:=\:\mathrm{2}{h}−\pi/\mathrm{2} \\ $$$${I}=\mathrm{2}\sqrt{\mathrm{2}}\:\int\:\frac{\mathrm{sin}\:{h}}{\mathrm{2}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{h}}\:{dh} \\ $$$${letting}\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:{h}\:=\:{u}\: \\ $$$${I}=−\mathrm{2}\int\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\:−\mathrm{2arctan}\:\left({u}\right)+\:{c} \\ $$$$=\:−\mathrm{2arctan}\:\left(\sqrt{\mathrm{2}}\:\mathrm{cos}\:\left({x}+\pi/\mathrm{4}\right)\right)+\:{c}\: \\ $$

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)))

$$\left.\mathrm{1}\right)\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{ln2}} \:\mathrm{x}^{−\mathrm{2}} \:\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \:\mathrm{ex}\:=\left[\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{x}}} \right]_{\mathrm{0}} ^{\mathrm{ln2}} \:=\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{ln2}}} −\mathrm{0}\:=\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{ln2}}} \\ $$

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