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Question Number 179194 by Acem last updated on 26/Oct/22
Evaluate the ∫ ((tan^5 x)/(cos^9 x)) dx
$${Evaluate}\:{the}\:\int\:\frac{\mathrm{tan}^{\mathrm{5}} \:{x}}{\mathrm{cos}^{\mathrm{9}} \:{x}}\:{dx} \\ $$
Answered by Acem last updated on 26/Oct/22
Deuxie′me methode: ∫ ((tan^5 x)/(cos^9 x)) dx= ∫sec^9 x tan^5 x dx Par remplacement u= sec x , du= sec x tan x dx I= ∫u^8 (u^2 −1)^2 du ; tan^2 x= sec^2 x −1 I= (u^(13) /(13)) −((2u^(11) )/(11)) + (u^9 /9) I= (1/(13)) sec^(13) x −(2/(11)) sec^(11) x +(1/9) sec^9 x + c
$${Deuxie}'{me}\:{methode}: \\ $$$$\int\:\frac{\mathrm{tan}^{\mathrm{5}} \:{x}}{\mathrm{cos}^{\mathrm{9}} \:{x}}\:{dx}=\:\int\mathrm{sec}^{\mathrm{9}} \:{x}\:\mathrm{tan}^{\mathrm{5}} \:{x}\:{dx} \\ $$$$\:{Par}\:{remplacement} \\ $$$$\:{u}=\:\mathrm{sec}\:\:{x}\:,\:{du}=\:\mathrm{sec}\:{x}\:\mathrm{tan}\:{x}\:{dx} \\ $$$$\: \\ $$$$\:{I}=\:\int{u}^{\mathrm{8}} \:\left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} {du}\:;\:\mathrm{tan}^{\mathrm{2}} \:{x}=\:\mathrm{sec}^{\mathrm{2}} \:{x}\:−\mathrm{1} \\ $$$$\:{I}=\:\frac{{u}^{\mathrm{13}} }{\mathrm{13}}\:−\frac{\mathrm{2}{u}^{\mathrm{11}} }{\mathrm{11}}\:+\:\frac{{u}^{\mathrm{9}} }{\mathrm{9}} \\ $$$$\:{I}=\:\frac{\mathrm{1}}{\mathrm{13}}\:\mathrm{sec}^{\mathrm{13}} \:{x}\:−\frac{\mathrm{2}}{\mathrm{11}}\:\mathrm{sec}^{\mathrm{11}} \:{x}\:+\frac{\mathrm{1}}{\mathrm{9}}\:\mathrm{sec}^{\mathrm{9}} \:{x}\:+\:{c} \\ $$
Answered by CElcedricjunior last updated on 30/Oct/22
∫((tan^5 x)/(cos^9 x))dx=∫((sin^5 x)/(cos^(14) x))d1x=∫sin^4 x×((sinx)/(cos^(14) x))dx posons { ((u=si^4 nx)),((v′=((sinx)/(cos^(14) x)))) :}=> { ((u′=4cosxsin^3 x)),((v=(1/(13cos^(13) x)))) :} =((sin^4 x)/(13cos^(13) x))−(4/(13))∫((sin^3 x)/(cos^(12) x))dx posons { ((u=sin^2 x)),((v′=((sinx)/(cos^(12) x)))) :}=> { ((u′=2sinxcosx)),((v=(1/(11cos^(11) x)))) :} =((tan^4 x)/(13cos^9 ))−((4tan^2 x)/(143cos^9 ))+(8/(143))∫((sinx)/(cos^(10) x))dx =((tan^4 x)/(13cos^9 ))−((4ta^2 nx)/(143cos^9 x))+(8/(143))((1/(9cos^9 x))) d′ou^� ∫((tan^5 x)/(cos^9 x))dx=((99tan^4 x−36tan^2 x+8)/(1287cos^9 x)) +k ou^� k∈R ..................le celebre cedric junior...........
$$\int\frac{\boldsymbol{{ta}}\overset{\mathrm{5}} {\boldsymbol{{n}x}}}{\boldsymbol{{cos}}^{\mathrm{9}} \boldsymbol{{x}}}\boldsymbol{{dx}}=\int\frac{\boldsymbol{{si}}\overset{\mathrm{5}} {\boldsymbol{{n}x}}}{\boldsymbol{{c}}{o}\overset{\mathrm{14}} {{s}x}}{d}\mathrm{1}{x}=\int\boldsymbol{\mathrm{si}}\overset{\mathrm{4}} {\boldsymbol{\mathrm{n}x}}×\frac{\boldsymbol{\mathrm{sinx}}}{\boldsymbol{\mathrm{co}}\overset{\mathrm{14}} {\boldsymbol{\mathrm{s}x}}}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{posons}}\: \\ $$$$\begin{cases}{\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{s}}\overset{\mathrm{4}} {\boldsymbol{\mathrm{i}nx}}}\\{\boldsymbol{\mathrm{v}}'=\frac{\boldsymbol{\mathrm{sinx}}}{\boldsymbol{\mathrm{co}}\overset{\mathrm{14}} {\boldsymbol{\mathrm{s}x}}}}\end{cases}=>\begin{cases}{\boldsymbol{\mathrm{u}}'=\mathrm{4}\boldsymbol{\mathrm{cosxsi}}\overset{\mathrm{3}} {\boldsymbol{\mathrm{n}x}}}\\{\boldsymbol{\mathrm{v}}=\frac{\mathrm{1}}{\mathrm{13}\boldsymbol{\mathrm{co}}\overset{\mathrm{13}} {\boldsymbol{\mathrm{s}x}}}}\end{cases} \\ $$$$=\frac{\boldsymbol{\mathrm{si}}\overset{\mathrm{4}} {\boldsymbol{\mathrm{n}x}}}{\mathrm{13}\boldsymbol{\mathrm{co}}\overset{\mathrm{13}} {\boldsymbol{\mathrm{s}x}}}−\frac{\mathrm{4}}{\mathrm{13}}\int\frac{\boldsymbol{\mathrm{si}}\overset{\mathrm{3}} {\boldsymbol{\mathrm{n}x}}}{\boldsymbol{\mathrm{co}}\overset{\mathrm{12}} {\boldsymbol{\mathrm{s}x}}}\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{posons}}\:\begin{cases}{\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{si}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}x}}}\\{\boldsymbol{\mathrm{v}}'=\frac{\boldsymbol{\mathrm{sinx}}}{\boldsymbol{\mathrm{co}}\overset{\mathrm{12}} {\boldsymbol{\mathrm{s}x}}}}\end{cases}=>\begin{cases}{\boldsymbol{\mathrm{u}}'=\mathrm{2}\boldsymbol{\mathrm{sinxcosx}}}\\{\boldsymbol{\mathrm{v}}=\frac{\mathrm{1}}{\mathrm{11}\boldsymbol{\mathrm{co}}\overset{\mathrm{11}} {\boldsymbol{\mathrm{s}x}}}}\end{cases} \\ $$$$=\frac{\boldsymbol{\mathrm{ta}}\overset{\mathrm{4}} {\boldsymbol{\mathrm{n}x}}}{\mathrm{13}\boldsymbol{\mathrm{co}}\overset{\mathrm{9}} {\boldsymbol{\mathrm{s}}}}−\frac{\mathrm{4}\boldsymbol{\mathrm{ta}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{n}x}}}{\mathrm{143}\boldsymbol{\mathrm{co}}\overset{\mathrm{9}} {\boldsymbol{\mathrm{s}}}}+\frac{\mathrm{8}}{\mathrm{143}}\int\frac{\boldsymbol{\mathrm{sinx}}}{\boldsymbol{\mathrm{co}}\overset{\mathrm{10}} {\boldsymbol{\mathrm{s}x}}}\boldsymbol{\mathrm{dx}} \\ $$$$=\frac{\boldsymbol{\mathrm{ta}}\overset{\mathrm{4}} {\boldsymbol{\mathrm{n}x}}}{\mathrm{13}\boldsymbol{\mathrm{cos}}^{\mathrm{9}} }−\frac{\mathrm{4}\boldsymbol{\mathrm{t}}\overset{\mathrm{2}} {\boldsymbol{\mathrm{a}nx}}}{\mathrm{143}\boldsymbol{\mathrm{co}}\overset{\mathrm{9}} {\boldsymbol{\mathrm{s}x}}}+\frac{\mathrm{8}}{\mathrm{143}}\left(\frac{\mathrm{1}}{\mathrm{9}\boldsymbol{\mathrm{co}}\overset{\mathrm{9}} {\boldsymbol{\mathrm{s}x}}}\right) \\ $$$$\boldsymbol{\mathrm{d}}'\boldsymbol{\mathrm{o}}\grave {\boldsymbol{\mathrm{u}}} \\ $$$$\int\frac{\boldsymbol{{ta}}\overset{\mathrm{5}} {\boldsymbol{{n}x}}}{\boldsymbol{{co}}\overset{\mathrm{9}} {\boldsymbol{{s}x}}}\boldsymbol{{dx}}=\frac{\mathrm{99}\boldsymbol{{ta}}\overset{\mathrm{4}} {\boldsymbol{{n}x}}−\mathrm{36}\boldsymbol{{ta}}\overset{\mathrm{2}} {\boldsymbol{{n}x}}+\mathrm{8}}{\mathrm{1287}\boldsymbol{\mathrm{co}}\overset{\mathrm{9}} {\boldsymbol{\mathrm{s}x}}}\:+\boldsymbol{{k}} \\ $$$$\boldsymbol{{o}}\grave {\boldsymbol{{u}}}\:\boldsymbol{\mathrm{k}}\in\mathbb{R} \\ $$$$\: \\ $$$$\:………………{le}\:{celebre}\:{cedric}\:{junior}……….. \\ $$
Commented by Acem last updated on 26/Oct/22
Good Sir!
$${Good}\:{Sir}! \\ $$

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