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Question Number 75669 by peter frank last updated on 15/Dec/19

Commented by mathmax by abdo last updated on 15/Dec/19

let A =∫ (dθ/(sin^4 θ +cos^4 θ)) ⇒ A =∫ (dθ/((cos^2 θ+sin^2 θ)^2 −2cos^2 θ sin^2 θ)) =∫ (dθ/(1−2((1/2)sin(2θ))^2 )) =∫ (dθ/(1−(1/2)sin^2 (2θ))) =∫ (dθ/(1−(1/2)(((1−cos(4θ))/2)))) =∫ ((4dθ)/(4−1+cos(4θ))) =4∫ (dθ/(3+cos(4θ))) =_(4θ =t) 4 ∫ (dt/(4(3+cost))) =∫ (dt/(3+cost)) =_(tan((t/2))=u) ∫ (1/(3+((1−u^2 )/(1+u^2 ))))((2du)/(1+u^2 )) =∫ ((2du)/(3+3u^2 +1−u^2 )) =∫ ((2du)/(4+2u^2 )) =∫ (du/(2+u^2 )) =_(u=(√2)α) ∫ (((√2)dα)/(2(1+α^2 ))) =((√2)/2) arctan(α)+C =((√2)/2) arctan((u/(√2)))+C =((√2)/2) arctan((1/(√2))tan((t/2)))+C =((√2)/2) arctan((1/(√2))tan(2θ)) +C .

$${let}\:{A}\:=\int\:\frac{{d}\theta}{{sin}^{\mathrm{4}} \theta\:+{cos}^{\mathrm{4}} \theta}\:\Rightarrow\:{A}\:=\int\:\:\:\:\frac{{d}\theta}{\left({cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} −\mathrm{2}{cos}^{\mathrm{2}} \theta\:{sin}^{\mathrm{2}} \theta} \\ $$$$=\int\:\:\frac{{d}\theta}{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}\theta\right)\right)^{\mathrm{2}} }\:=\int\:\:\frac{{d}\theta}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\mathrm{2}\theta\right)}\:=\int\:\:\frac{{d}\theta}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}\right)} \\ $$$$=\int\:\:\frac{\mathrm{4}{d}\theta}{\mathrm{4}−\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}\:=\mathrm{4}\int\:\:\frac{{d}\theta}{\mathrm{3}+{cos}\left(\mathrm{4}\theta\right)}\:=_{\mathrm{4}\theta\:={t}} \:\:\mathrm{4}\:\int\:\:\:\frac{{dt}}{\mathrm{4}\left(\mathrm{3}+{cost}\right)} \\ $$$$=\int\:\:\:\frac{{dt}}{\mathrm{3}+{cost}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\:\int\:\:\:\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{\mathrm{2}{du}}{\mathrm{3}+\mathrm{3}{u}^{\mathrm{2}} +\mathrm{1}−{u}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{du}}{\mathrm{4}+\mathrm{2}{u}^{\mathrm{2}} }\:=\int\:\:\:\frac{{du}}{\mathrm{2}+{u}^{\mathrm{2}} }\:=_{{u}=\sqrt{\mathrm{2}}\alpha} \:\:\:\int\:\:\:\frac{\sqrt{\mathrm{2}}{d}\alpha}{\mathrm{2}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left(\alpha\right)+{C}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left(\frac{{u}}{\sqrt{\mathrm{2}}}\right)+{C} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}\left(\frac{{t}}{\mathrm{2}}\right)\right)+{C}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}\left(\mathrm{2}\theta\right)\right)\:+{C}\:\:. \\ $$$$ \\ $$

Commented by peter frank last updated on 15/Dec/19

thank you

$${thank}\:{you} \\ $$

Commented by mathmax by abdo last updated on 15/Dec/19

you are welcome.

$${you}\:{are}\:{welcome}. \\ $$

Answered by mr W last updated on 15/Dec/19

sin^2 θ+cos^2 θ=1 (sin^2 θ+cos^2 θ)^2 =1^2 sin^4 θ+cos^4 θ+2 sin^2 θ cos^2 θ=1 sin^4 θ+cos^4 θ+((sin^2 2θ)/2)=1 sin^4 θ+cos^4 θ=((2−sin^2 2θ)/2) sin^4 θ+cos^4 θ=((((√2)−sin 2θ)((√2)+sin 2θ))/2) ∫(dθ/(sin^4 θ+cos^4 θ)) =∫((2 dθ)/(((√2)−sin 2θ)((√2)+sin 2θ))) =∫(du/(((√2)−sin u)((√2)+sin u))) with u=2θ =(1/(2(√2)))∫[(1/((√2)−sin u))+(1/((√2)+sin u))]du =(1/(2(√2)))[2 tan^(−1) ((√2) tan (u/2)−1)+2 tan^(−1) ((√2) tan (u/2)+1)]+C =(1/(√2))[tan^(−1) ((√2) tan θ−1)+tan^(−1) ((√2) tan θ+1)]+C

$$\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{cos}^{\mathrm{2}} \:\theta=\mathrm{1} \\ $$$$\left(\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{cos}^{\mathrm{2}} \:\theta\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{4}} \:\theta+\mathrm{cos}^{\mathrm{4}} \:\theta+\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta=\mathrm{1} \\ $$$$\mathrm{sin}^{\mathrm{4}} \:\theta+\mathrm{cos}^{\mathrm{4}} \:\theta+\frac{\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta}{\mathrm{2}}=\mathrm{1} \\ $$$$\mathrm{sin}^{\mathrm{4}} \:\theta+\mathrm{cos}^{\mathrm{4}} \:\theta=\frac{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta}{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{4}} \:\theta+\mathrm{cos}^{\mathrm{4}} \:\theta=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\mathrm{2}\theta\right)\left(\sqrt{\mathrm{2}}+\mathrm{sin}\:\mathrm{2}\theta\right)}{\mathrm{2}} \\ $$$$\int\frac{{d}\theta}{\mathrm{sin}^{\mathrm{4}} \:\theta+\mathrm{cos}^{\mathrm{4}} \:\theta} \\ $$$$=\int\frac{\mathrm{2}\:{d}\theta}{\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:\mathrm{2}\theta\right)\left(\sqrt{\mathrm{2}}+\mathrm{sin}\:\mathrm{2}\theta\right)} \\ $$$$=\int\frac{{du}}{\left(\sqrt{\mathrm{2}}−\mathrm{sin}\:{u}\right)\left(\sqrt{\mathrm{2}}+\mathrm{sin}\:{u}\right)}\:{with}\:{u}=\mathrm{2}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\left[\frac{\mathrm{1}}{\sqrt{\mathrm{2}}−\mathrm{sin}\:{u}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}+\mathrm{sin}\:{u}}\right]{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\:\mathrm{tan}\:\frac{{u}}{\mathrm{2}}−\mathrm{1}\right)+\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\:\mathrm{tan}\:\frac{{u}}{\mathrm{2}}+\mathrm{1}\right)\right]+{C} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\:\mathrm{tan}\:\theta−\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\:\mathrm{tan}\:\theta+\mathrm{1}\right)\right]+{C} \\ $$

Commented by peter frank last updated on 15/Dec/19

thank you

$${thank}\:{you} \\ $$

Commented by behi83417@gmail.com last updated on 15/Dec/19

=(1/(√2)).tg^(−1) ((((√2)tgθ)/(1−tg^2 θ)))+C=(1/(√2))tg^(−1) (((√2)/2).tg2𝛉)+C

$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}.\mathrm{tg}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\mathrm{tg}\theta}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \theta}\right)+\mathrm{C}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{tg}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}.\boldsymbol{\mathrm{tg}}\mathrm{2}\boldsymbol{\theta}\right)+\boldsymbol{\mathrm{C}} \\ $$

Answered by vishalbhardwaj last updated on 24/Dec/19

∫ ((sec^4 θ dθ)/(tan^2 θ+1)) = ∫ ((sec^2 θ.sec^2 θ)/(tan^4 θ+1)) dθ ∫ (((1+tan^2 θ)sec^2 θ)/((1+tan^4 θ))) dθ (let tanθ =t) ⇒ sec^2 θ dθ = dt = ∫ (((1+t^2 ))/((1+t^4 ))) dt = ∫ ((t^2 (1+(1/t^2 )))/(t^2 (t^2 +(1/t^2 )))) dt = ∫ (((1+(1/t^2 )))/((t^2 +(1/t^2 )))) dt = ∫ (((1+(1/t^2 )))/((t−(1/t))^2 +((√2))^2 )) dt let t−(1/t)= z ⇒ (1+(1/t^2 )) dt = dz = ∫ (dz/(z^2 +((√2))^2 )) = (1/(√2)) tan^(−1) ((z/(√2))) +C = (1/(√2)) tan^(−1) (((t−(1/t))/(√2))) +C = (1/(√2)) tan^(−1) (((tanx−cotx)/(√2)))+C

$$\int\:\:\frac{{sec}^{\mathrm{4}} \theta\:{d}\theta}{{tan}^{\mathrm{2}} \theta+\mathrm{1}}\:=\:\int\:\frac{{sec}^{\mathrm{2}} \theta.{sec}^{\mathrm{2}} \theta}{{tan}^{\mathrm{4}} \theta+\mathrm{1}}\:{d}\theta\: \\ $$$$\int\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){sec}^{\mathrm{2}} \theta}{\left(\mathrm{1}+{tan}^{\mathrm{4}} \theta\right)}\:{d}\theta\:\:\left({let}\:{tan}\theta\:=\mathrm{t}\right) \\ $$$$\Rightarrow\:\:{sec}^{\mathrm{2}} \theta\:{d}\theta\:=\:{dt}\: \\ $$$$=\:\int\:\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}\:{dt} \\ $$$$=\:\int\:\frac{{t}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}\:{dt} \\ $$$$=\:\int\:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}\:{dt} \\ $$$$=\:\int\:\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:{dt} \\ $$$${let}\:{t}−\frac{\mathrm{1}}{{t}}=\:{z}\:\Rightarrow\:\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)\:{dt}\:=\:{dz} \\ $$$$=\:\int\:\frac{{dz}}{{z}^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{tan}^{−\mathrm{1}} \left(\frac{{z}}{\sqrt{\mathrm{2}}}\right)\:+{C} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\sqrt{\mathrm{2}}}\right)\:+{C} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{tan}^{−\mathrm{1}} \left(\frac{{tanx}−{cotx}}{\sqrt{\mathrm{2}}}\right)+{C} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by malwaan last updated on 15/Dec/19

fantastic thankx 4 all

$$\boldsymbol{{fantastic}} \\ $$$$\boldsymbol{{thankx}}\:\mathrm{4}\:\boldsymbol{{all}} \\ $$

Answered by MJS last updated on 15/Dec/19

∫(dθ/(sin^4 θ +cos^4 θ))=4∫(dθ/(3+cos 4θ))= [t=(1/2)tan 2θ → dθ=(dt/(2(t^2 +1)))] =∫(dt/(t^2 +2))=((√2)/2)arctan ((t(√2))/2) = =((√2)/2)arctan (((√2)tan 2θ)/2) +C

$$\int\frac{{d}\theta}{\mathrm{sin}^{\mathrm{4}} \:\theta\:+\mathrm{cos}^{\mathrm{4}} \:\theta}=\mathrm{4}\int\frac{{d}\theta}{\mathrm{3}+\mathrm{cos}\:\mathrm{4}\theta}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\mathrm{2}\theta\:\rightarrow\:{d}\theta=\frac{{dt}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\right] \\ $$$$=\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\frac{{t}\sqrt{\mathrm{2}}}{\mathrm{2}}\:= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\mathrm{tan}\:\mathrm{2}\theta}{\mathrm{2}}\:+{C} \\ $$

Commented by peter frank last updated on 15/Dec/19

thank you

$${thank}\:{you} \\ $$

Commented by peter frank last updated on 02/Jan/20

sorry sir i pressing flag option by mistake

$${sorry}\:{sir}\:{i}\:{pressing} \\ $$$${flag}\:{option}\:{by}\:{mistake} \\ $$

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