t-5-2-t-2-dt-

Question Number 120040 by bramlexs22 last updated on 28/Oct/20

$$\:\int\:\frac{{t}^{\mathrm{5}} }{\:\sqrt{\mathrm{2}+{t}^{\mathrm{2}} }}\:{dt}\: \\ $$

Answered by john santu last updated on 28/Oct/20

let 2+t^2 = h^2 → t dt = h dh ∫ ((t^4 tdt)/( (√(2+t^2 )))) = ∫ (((h^2 −2)^2 h dh)/h) ∫ (h^4 −4h^2 +4) dh = (1/5)h^5 −(4/3)h^3 +4h + c =(1/(15))(√(2+t^2 )) {3(2+t^2 )^2 −20(2+t^2 )+60 }+ c = ((√(2+t^2 ))/(15)) { 3t^4 −8t^2 +32 } + c

$$\:{let}\:\mathrm{2}+{t}^{\mathrm{2}} \:=\:{h}^{\mathrm{2}} \:\rightarrow\:{t}\:{dt}\:=\:{h}\:{dh}\: \\ $$$$\int\:\frac{{t}^{\mathrm{4}} \:{tdt}}{\:\sqrt{\mathrm{2}+{t}^{\mathrm{2}} }}\:=\:\int\:\frac{\left({h}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \:{h}\:{dh}}{{h}} \\ $$$$\int\:\left({h}^{\mathrm{4}} −\mathrm{4}{h}^{\mathrm{2}} +\mathrm{4}\right)\:{dh}\:=\:\frac{\mathrm{1}}{\mathrm{5}}{h}^{\mathrm{5}} −\frac{\mathrm{4}}{\mathrm{3}}{h}^{\mathrm{3}} +\mathrm{4}{h}\:+\:{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{15}}\sqrt{\mathrm{2}+{t}^{\mathrm{2}} }\:\left\{\mathrm{3}\left(\mathrm{2}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{20}\left(\mathrm{2}+{t}^{\mathrm{2}} \right)+\mathrm{60}\:\right\}+\:{c} \\ $$$$=\:\frac{\sqrt{\mathrm{2}+{t}^{\mathrm{2}} }}{\mathrm{15}}\:\left\{\:\mathrm{3}{t}^{\mathrm{4}} −\mathrm{8}{t}^{\mathrm{2}} +\mathrm{32}\:\right\}\:+\:{c}\: \\ $$

Commented by Lordose last updated on 28/Oct/20

FASTEST

Commented by bramlexs22 last updated on 29/Oct/20

$${waw}… \\ $$

Answered by Lordose last updated on 28/Oct/20

set t= (√2)tanx ⇒ dt = (√2)(1+t^2 )dx ∫ ((((√2))^6 tan^5 xsec^2 x)/( (√2)secx))dx 4(√2)∫ tan^5 xsecxdx 4(√2)∫tanxsecx(sec^2 x−1)^2 dx u=secx ⇒ du= secxtanxdx 4(√2)∫(u^2 −1)^2 du 4(√2)∫(u^4 −2u^2 +1)du 4(√2)((u^5 /5) − ((2u^3 )/3) + u) + C 4(√2)(((sec^5 x)/5) − ((2sec^3 (x))/3) + secx) + C x = tan^(−1) ((t/( (√2)))) I = (1/(15))(√(t^2 +2))(3t^2 −8t+32) + C

$$\mathrm{set}\:\mathrm{t}=\:\sqrt{\mathrm{2}}\mathrm{tanx}\:\Rightarrow\:\mathrm{dt}\:=\:\sqrt{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\int\:\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{6}} \mathrm{tan}^{\mathrm{5}} \mathrm{xsec}^{\mathrm{2}} \mathrm{x}}{\:\sqrt{\mathrm{2}}\mathrm{secx}}\mathrm{dx} \\ $$$$\mathrm{4}\sqrt{\mathrm{2}}\int\:\mathrm{tan}^{\mathrm{5}} \mathrm{xsecxdx} \\ $$$$\mathrm{4}\sqrt{\mathrm{2}}\int\mathrm{tanxsecx}\left(\mathrm{sec}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{secx}\:\Rightarrow\:\mathrm{du}=\:\mathrm{secxtanxdx} \\ $$$$\mathrm{4}\sqrt{\mathrm{2}}\int\left(\mathrm{u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \mathrm{du} \\ $$$$\mathrm{4}\sqrt{\mathrm{2}}\int\left(\mathrm{u}^{\mathrm{4}} −\mathrm{2u}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{du} \\ $$$$\mathrm{4}\sqrt{\mathrm{2}}\left(\frac{\mathrm{u}^{\mathrm{5}} }{\mathrm{5}}\:−\:\frac{\mathrm{2u}^{\mathrm{3}} }{\mathrm{3}}\:+\:\mathrm{u}\right)\:+\:\mathrm{C} \\ $$$$\mathrm{4}\sqrt{\mathrm{2}}\left(\frac{\mathrm{sec}^{\mathrm{5}} \mathrm{x}}{\mathrm{5}}\:−\:\frac{\mathrm{2sec}^{\mathrm{3}} \left(\mathrm{x}\right)}{\mathrm{3}}\:+\:\mathrm{secx}\right)\:+\:\mathrm{C} \\ $$$$\mathrm{x}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{t}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{15}}\sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{2}}\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{8t}+\mathrm{32}\right)\:+\:\mathrm{C} \\ $$$$ \\ $$

Commented by bramlexs22 last updated on 29/Oct/20

$${yes} \\ $$

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