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1-dx-1-sin-x-R-solve-in-2-4-x-4-x-4-82-

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Question Number 62754 by aliesam last updated on 24/Jun/19
1)∫(dx/(1−sin(x))) R solve in(2) (4−x)^4 +x^4 =82
$$\left.\mathrm{1}\right)\int\frac{{dx}}{\mathrm{1}−{sin}\left({x}\right)} \\ $$$$ \\ $$$${R}\:{solve}\:{in}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{4}−{x}\right)^{\mathrm{4}} +{x}^{\mathrm{4}} =\mathrm{82} \\ $$
Commented by Tony Lin last updated on 25/Jun/19
(2)(4−x)^4 +x^4 =81+1=3^4 +1^4 f(x)=(4−x)^4 +x^4 −82 f(1)=0, f(3)=0 ⇒f(x)=2(x^4 −8x^3 +48x^2 −128x+87) =2(x−1)(x−3)(x^2 −4x+29) △=(−4)^2 −4×29<0 x^2 −4x+29 has no roots in R ⇒x=1 or x=3
$$\left(\mathrm{2}\right)\left(\mathrm{4}−{x}\right)^{\mathrm{4}} +{x}^{\mathrm{4}} =\mathrm{81}+\mathrm{1}=\mathrm{3}^{\mathrm{4}} +\mathrm{1}^{\mathrm{4}} \\ $$$${f}\left({x}\right)=\left(\mathrm{4}−{x}\right)^{\mathrm{4}} +{x}^{\mathrm{4}} −\mathrm{82} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0},\:{f}\left(\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{2}\left({x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{3}} +\mathrm{48}{x}^{\mathrm{2}} −\mathrm{128}{x}+\mathrm{87}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{29}\right) \\ $$$$\bigtriangleup=\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{29}<\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{29}\:{has}\:{no}\:{roots}\:{in}\:{R} \\ $$$$\Rightarrow{x}=\mathrm{1}\:{or}\:{x}=\mathrm{3} \\ $$
Commented by mathmax by abdo last updated on 25/Jun/19
let use another way let I =∫ (dx/(1−sinx)) changement tan((x/2))=t give I =∫ ((2dt)/((1+t^2 )(1−((2t)/(1+t^2 ))))) =∫ ((2dt)/(1+t^2 −2t)) =∫ ((2dt)/((t−1)^2 )) =−(2/(t−1)) +c =(2/(1−tan((x/2)))) +c .
$${let}\:{use}\:{another}\:{way}\:{let}\:{I}\:=\int\:\:\frac{{dx}}{\mathrm{1}−{sinx}}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\int\:\:\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)}\:=\int\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{2}{t}}\:=\int\:\:\frac{\mathrm{2}{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{2}}{{t}−\mathrm{1}}\:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\:+{c}\:. \\ $$
Answered by Hope last updated on 25/Jun/19
1)∫((1+sinx)/(cos^2 x))dx ∫sec^2 x+secxtanx dx tanx+secx+c
$$\left.\mathrm{1}\right)\int\frac{\mathrm{1}+{sinx}}{{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$\int{sec}^{\mathrm{2}} {x}+{secxtanx}\:\:{dx} \\ $$$${tanx}+{secx}+{c} \\ $$

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