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Question Number 179244 by Acem last updated on 27/Oct/22
Evaluate ∫tan^4 x sec^5 x dx
$${Evaluate}\:\int\mathrm{tan}^{\mathrm{4}} \:{x}\:\mathrm{sec}^{\mathrm{5}} \:{x}\:{dx} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 27/Oct/22
I=∫tan^4 xsec^5 xdx=∫tan^3 xsec^4 x∙secxtanxdx =(1/5)tan^3 xsec^5 x−(3/5)∫tan^2 xsec^7 xdx =(1/5)tan^3 xsec^5 x−(3/(35))tanxsec^7 x+(3/(35))∫sec^9 xdx =(1/5)tan^3 xsec^5 x−(3/5)tanxsec^7 x+(3/(35))∫((cosx)/((1−sin^2 x)^5 ))dx (1/((1−t^2 )^5 ))=(a/(t+1))+(b/((t+1)^2 ))+(c/((t+1)^3 ))+(d/((t+1)^4 ))+(e/((t+1)^5 )) +(f/(t−1))+(g/((t−1)^2 ))+(h/((t−1)^3 ))+(i/((t−1)^4 ))+(j/((t−1)^5 )) Continuing this process we′ll be able to decompose (1/((1−t^2 )^5 )) completely and integrate (1/((1−t^2 )^5 )) , where t=sinx
$${I}=\int\mathrm{tan}^{\mathrm{4}} {x}\mathrm{sec}^{\mathrm{5}} {xdx}=\int\mathrm{tan}^{\mathrm{3}} {x}\mathrm{sec}^{\mathrm{4}} {x}\centerdot\mathrm{sec}{x}\mathrm{tan}{xdx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{tan}^{\mathrm{3}} {x}\mathrm{sec}^{\mathrm{5}} {x}−\frac{\mathrm{3}}{\mathrm{5}}\int\mathrm{tan}^{\mathrm{2}} {x}\mathrm{sec}^{\mathrm{7}} {xdx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{tan}^{\mathrm{3}} {x}\mathrm{sec}^{\mathrm{5}} {x}−\frac{\mathrm{3}}{\mathrm{35}}\mathrm{tan}{x}\mathrm{sec}^{\mathrm{7}} {x}+\frac{\mathrm{3}}{\mathrm{35}}\int\mathrm{sec}^{\mathrm{9}} {xdx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{tan}^{\mathrm{3}} {x}\mathrm{sec}^{\mathrm{5}} {x}−\frac{\mathrm{3}}{\mathrm{5}}\mathrm{tan}{x}\mathrm{sec}^{\mathrm{7}} {x}+\frac{\mathrm{3}}{\mathrm{35}}\int\frac{\mathrm{cos}{x}}{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\right)^{\mathrm{5}} }{dx} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{5}} }=\frac{{a}}{{t}+\mathrm{1}}+\frac{{b}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{c}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }+\frac{{d}}{\left({t}+\mathrm{1}\right)^{\mathrm{4}} }+\frac{{e}}{\left({t}+\mathrm{1}\right)^{\mathrm{5}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{f}}{{t}−\mathrm{1}}+\frac{{g}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{h}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{{i}}{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }+\frac{{j}}{\left({t}−\mathrm{1}\right)^{\mathrm{5}} } \\ $$$$\mathrm{Continuing}\:\mathrm{this}\:\mathrm{process}\:\mathrm{we}'\mathrm{ll}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{decompose}\: \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{5}} }\:\mathrm{completely}\:\mathrm{and}\:\mathrm{integrate}\:\frac{\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{5}} }\:,\:\mathrm{where}\:{t}=\mathrm{sin}{x} \\ $$
Commented by ARUNG_Brandon_MBU last updated on 27/Oct/22
(1/((1−t^2 )^5 ))=((35)/(256(1+t)))+((35)/(256(1+t)^2 ))+((15)/(128(1+t)^3 ))+(5/(64(1+t)^4 ))+(1/(32(1+t)^5 )) −((35)/(256(1+t)))+((35)/(256(1+t)^2 ))−((15)/(128(1+t)^3 ))+(5/(64(1+t)^4 ))−(1/(32(1+t)^5 ))
$$\frac{\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{5}} }=\frac{\mathrm{35}}{\mathrm{256}\left(\mathrm{1}+{t}\right)}+\frac{\mathrm{35}}{\mathrm{256}\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }+\frac{\mathrm{15}}{\mathrm{128}\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{64}\left(\mathrm{1}+{t}\right)^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{32}\left(\mathrm{1}+{t}\right)^{\mathrm{5}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{35}}{\mathrm{256}\left(\mathrm{1}+{t}\right)}+\frac{\mathrm{35}}{\mathrm{256}\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }−\frac{\mathrm{15}}{\mathrm{128}\left(\mathrm{1}+{t}\right)^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{64}\left(\mathrm{1}+{t}\right)^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{32}\left(\mathrm{1}+{t}\right)^{\mathrm{5}} } \\ $$
Answered by MJS_new last updated on 27/Oct/22
∫tan^4 x sec^5 x dx=∫((sin^4 x)/(cos^9 x))dx= [t=sin x → dx=(dt/(cos x))] =−∫(t^4 /((t^2 −1)^5 ))dt= [Ostrogradski] =−((t(t^2 +1)(3t^4 −14t^2 +3))/(128(t^2 −1)^4 ))−(3/(128))∫(dt/(t^2 −1))= =−((t(t^2 +1)(3t^4 −14t^2 +3))/(128(t^2 −1)^4 ))+(3/(128))ln ∣((1+t)/(1−t))∣ =...
$$\int\mathrm{tan}^{\mathrm{4}} \:{x}\:\mathrm{sec}^{\mathrm{5}} \:{x}\:{dx}=\int\frac{\mathrm{sin}^{\mathrm{4}} \:{x}}{\mathrm{cos}^{\mathrm{9}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sin}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{cos}\:{x}}\right] \\ $$$$=−\int\frac{{t}^{\mathrm{4}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{5}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=−\frac{{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{t}^{\mathrm{4}} −\mathrm{14}{t}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{128}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }−\frac{\mathrm{3}}{\mathrm{128}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=−\frac{{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{t}^{\mathrm{4}} −\mathrm{14}{t}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{128}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{3}}{\mathrm{128}}\mathrm{ln}\:\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:=… \\ $$

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