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Question Number 35058 by math khazana by abdo last updated on 14/May/18

find ∫_0 ^(π/4) (dt/((1+cos^2 t)^3 ))

$${find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{cos}^{\mathrm{2}} {t}\right)^{\mathrm{3}} } \\ $$

Commented by abdo mathsup 649 cc last updated on 18/May/18

let put I = ∫_0 ^(π/4) (dt/((1+cos^2 t)^3 )) I = ∫_0 ^(π/4) (dt/((1+((1+cos(2t))/2))^3 )) = 8 ∫_0 ^(π/4) (dt/((3+cos(2t))^3 )) =_(2t=x) 8 ∫_0 ^(π/2) (1/((3+cosx)^3 ))(dx/2) = 4 ∫_0 ^(π/2) (dx/((3+cosx)^3 )) .changement tan((x/2))= t give I = 4 ∫_0 ^1 (1/((3 + ((1−t^2 )/(1+t^2 )))^3 )) ((2dt)/(1+t^2 )) I = 8 ∫_0 ^1 (((1+t^2 )^2 )/((3 +3t^2 +1−t^2 )^3 )) dt I = 8 ∫_0 ^1 (((1+t^2 )^2 )/((4 +2t^2 )^3 )) dt = ∫_0 ^1 (((1+t^2 )^2 )/((2+t^2 )^3 )) dt I = ∫_0 ^1 ((t^4 +2t^2 +1)/(8 +3.2^2 t^2 +3.2.t^4 +t^6 )) dt = ∫_0 ^1 ((t^4 +2t^2 +1)/(t^6 +6t^4 +12t^2 +8)) dt let decompose F(t) = ((t^4 +2t^2 +1)/(t^6 +6t^4 +12t^2 +8)).....be continued...

$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{cos}^{\mathrm{2}} {t}\right)^{\mathrm{3}} }\: \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$=\:\mathrm{8}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{{dt}}{\left(\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)\right)^{\mathrm{3}} }\:=_{\mathrm{2}{t}={x}} \:\:\:\mathrm{8}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}+{cosx}\right)^{\mathrm{3}} }\frac{{dx}}{\mathrm{2}} \\ $$$$=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{{dx}}{\left(\mathrm{3}+{cosx}\right)^{\mathrm{3}} }\:\:.{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)=\:{t} \\ $$$${give}\:\:{I}\:=\:\:\mathrm{4}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}\:+\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{3}} }\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${I}\:\:=\:\mathrm{8}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{3}\:+\mathrm{3}{t}^{\mathrm{2}} \:\:+\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{dt} \\ $$$${I}\:\:=\:\mathrm{8}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{4}\:+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{2}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{dt} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{t}^{\mathrm{4}} \:\:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{8}\:\:+\mathrm{3}.\mathrm{2}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{3}.\mathrm{2}.{t}^{\mathrm{4}} \:+{t}^{\mathrm{6}} }\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{6}} \:\:+\mathrm{6}{t}^{\mathrm{4}} \:\:+\mathrm{12}{t}^{\mathrm{2}} \:\:+\mathrm{8}}\:{dt}\:\:\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\:\frac{{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}}{{t}^{\mathrm{6}} \:+\mathrm{6}{t}^{\mathrm{4}} \:+\mathrm{12}{t}^{\mathrm{2}} \:+\mathrm{8}}.....{be}\:{continued}... \\ $$

Answered by MJS last updated on 16/May/18

∫(dt/((1+cos^2 (t))^3 ))= [cos(t)=(1/(sec(t))); sec^2 (t)=tan^2 (t)+1] =∫((sec^2 (t)(tan^2 (t)+1)^2 )/((tan^2 (t)+2)^3 ))dt= [u=tan(t) → dt=(du/(sec^2 (t)))] =∫(((u^2 +1)^2 )/((u^2 +2)^3 ))du=∫(du/(u^2 +2))−2∫(du/((u^2 +2)^2 ))+∫(du/((u^2 +2)^3 ))= ∫(du/(u^2 +2))= [v=((√2)/2)u → du=dv(√2)] =((√2)/2)∫(dv/(v^2 +1))=((√2)/2)arctan(v)=((√2)/2)arctan(((√2)/2)u) −2∫(du/((u^2 +2)^2 ))= [∫(du/((au^2 +b)^n ))=(u/(2b(n−1)(au^2 +b)^(n−1) ))+((2n−3)/(2b(n−1)))∫(du/((au^2 +b)^(n−1) ))] =−2((u/(4(u^2 +2)))+(1/4)∫(du/(u^2 +2)))= =−(u/(2(u^2 +2)))−((√2)/4)arctan(((√2)/2)u) ∫(du/((u^2 +2)^3 ))=(u/(8(u^2 +2)^2 ))+(3/8)∫(du/((u^2 +2)^2 ))= =(u/(8(u^2 +2)^2 ))+(3/8)((u/(4(u^2 +2)))+((√2)/8)arctan(((√2)/2)u))= =(u/(8(u^2 +2)^2 ))+((3u)/(32(u^2 +2)))+((3(√2))/(64))arctan(((√2)/2)u) =((19(√2))/(64))arctan(((√2)/2)u)−((13u)/(32(u^2 +2)))+(u/(8(u^2 +2)^2 ))= =((19(√2))/(64))arctan(((√2)/2)tan(t))−((tan(t)(13tan^2 (t)+22))/(32(tan^2 (t)+2)^2 ))+C ∫_0 ^(π/4) (dt/((1+cos^2 (t))^3 ))=((19(√2))/(64))arctan(((√2)/2))−((35)/(288))

$$\int\frac{{dt}}{\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left({t}\right)\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{cos}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{sec}\left({t}\right)};\:\mathrm{sec}^{\mathrm{2}} \left({t}\right)=\mathrm{tan}^{\mathrm{2}} \left({t}\right)+\mathrm{1}\right] \\ $$$$=\int\frac{\mathrm{sec}^{\mathrm{2}} \left({t}\right)\left(\mathrm{tan}^{\mathrm{2}} \left({t}\right)+\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{tan}^{\mathrm{2}} \left({t}\right)+\mathrm{2}\right)^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=\mathrm{tan}\left({t}\right)\:\rightarrow\:{dt}=\frac{{du}}{\mathrm{sec}^{\mathrm{2}} \left({t}\right)}\right] \\ $$$$=\int\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} }{du}=\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{2}}−\mathrm{2}\int\frac{{du}}{\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }+\int\frac{{du}}{\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} }= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{2}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{v}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\:\rightarrow\:{du}={dv}\sqrt{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{1}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\left({v}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\int\frac{{du}}{\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\int\frac{{du}}{\left({au}^{\mathrm{2}} +{b}\right)^{{n}} }=\frac{{u}}{\mathrm{2}{b}\left({n}−\mathrm{1}\right)\left({au}^{\mathrm{2}} +{b}\right)^{{n}−\mathrm{1}} }+\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{b}\left({n}−\mathrm{1}\right)}\int\frac{{du}}{\left({au}^{\mathrm{2}} +{b}\right)^{{n}−\mathrm{1}} }\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\mathrm{2}\left(\frac{{u}}{\mathrm{4}\left({u}^{\mathrm{2}} +\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{2}}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{{u}}{\mathrm{2}\left({u}^{\mathrm{2}} +\mathrm{2}\right)}−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{du}}{\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{3}} }=\frac{{u}}{\mathrm{8}\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{8}}\int\frac{{du}}{\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{{u}}{\mathrm{8}\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{8}}\left(\frac{{u}}{\mathrm{4}\left({u}^{\mathrm{2}} +\mathrm{2}\right)}+\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\right)\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{{u}}{\mathrm{8}\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }+\frac{\mathrm{3}{u}}{\mathrm{32}\left({u}^{\mathrm{2}} +\mathrm{2}\right)}+\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{64}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\right) \\ $$$$ \\ $$$$=\frac{\mathrm{19}\sqrt{\mathrm{2}}}{\mathrm{64}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\right)−\frac{\mathrm{13}{u}}{\mathrm{32}\left({u}^{\mathrm{2}} +\mathrm{2}\right)}+\frac{{u}}{\mathrm{8}\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{19}\sqrt{\mathrm{2}}}{\mathrm{64}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}\left({t}\right)\right)−\frac{\mathrm{tan}\left({t}\right)\left(\mathrm{13tan}^{\mathrm{2}} \left({t}\right)+\mathrm{22}\right)}{\mathrm{32}\left(\mathrm{tan}^{\mathrm{2}} \left({t}\right)+\mathrm{2}\right)^{\mathrm{2}} }+{C} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\frac{{dt}}{\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left({t}\right)\right)^{\mathrm{3}} }=\frac{\mathrm{19}\sqrt{\mathrm{2}}}{\mathrm{64}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)−\frac{\mathrm{35}}{\mathrm{288}} \\ $$

Commented by abdo mathsup 649 cc last updated on 15/May/18

thank you sir Mjs for this hard work...

$${thank}\:{you}\:{sir}\:{Mjs}\:{for}\:{this}\:{hard}\:{work}... \\ $$

Commented by MJS last updated on 16/May/18

you′re welcome, it keeps my brains young

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome},\:\mathrm{it}\:\mathrm{keeps}\:\mathrm{my}\:\mathrm{brains}\:\mathrm{young} \\ $$

Commented by abdo mathsup 649 cc last updated on 16/May/18

are you a doctor or enjener sir Mjs...

$${are}\:{you}\:{a}\:{doctor}\:{or}\:{enjener}\:{sir}\:{Mjs}... \\ $$

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