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Question Number 36659 by rahul 19 last updated on 03/Jun/18

∫ (1/(sin^4 x+cos^4 x)) dx

$$\int\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}\:{dx} \\ $$

Commented by MJS last updated on 03/Jun/18

see my answers to Qu. 36428

$$\mathrm{see}\:\mathrm{my}\:\mathrm{answers}\:\mathrm{to}\:\mathrm{Qu}.\:\mathrm{36428} \\ $$

Commented by prof Abdo imad last updated on 03/Jun/18

this integral is solved see the platform

$${this}\:{integral}\:{is}\:{solved}\:{see}\:{the}\:{platform} \\ $$

Commented by math khazana by abdo last updated on 11/Aug/18

let I = ∫ (dx/(sin^4 x +cos^4 x)) we have I = ∫ (dx/((cos^2 x +sin^2 x)^2 −2cos^2 x sin^2 x)) = ∫ (dx/(1−(1/2)sin^2 (2x))) =_(2x=t) ∫ (1/(1−(1/2)sin^2 t)) (dt/2) = ∫ (dt/(2−sin^2 t)) = ∫ (dt/(1+cos^2 t)) = ∫ (dt/(1+(1/(1+tan^2 t)))) = ∫ ((1+tan^2 t)/(2+tan^2 t)) dt =_(tant =u) ∫ ((1+u^2 )/(2+u^2 )) (du/(1+u^2 )) = ∫ (du/(2+u^2 )) =_(u=(√2)α) ∫ (((√2)dα)/(2(1+α^2 ))) =((√2)/2) arctan((u/(√2)))+c =((√2)/2) arctan(((tant)/(√2)))+c =((√2)/2) arctan(((tan(2x))/(√2))) +c ⇒ I =((√2)/2) arctan(((tan(2x))/(√2))) +c .

$${let}\:{I}\:\:=\:\int\:\:\:\frac{{dx}}{{sin}^{\mathrm{4}} {x}\:+{cos}^{\mathrm{4}} {x}}\:\:{we}\:{have}\: \\ $$$${I}\:\:=\:\int\:\:\:\:\:\frac{{dx}}{\left({cos}^{\mathrm{2}} {x}\:+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} \:−\mathrm{2}{cos}^{\mathrm{2}} {x}\:{sin}^{\mathrm{2}} {x}} \\ $$$$=\:\int\:\:\:\:\frac{{dx}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}\:=_{\mathrm{2}{x}={t}} \:\:\:\:\int\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} {t}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\int\:\:\:\:\:\frac{{dt}}{\mathrm{2}−{sin}^{\mathrm{2}} {t}}\:=\:\int\:\:\:\:\:\frac{{dt}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:=\:\int\:\:\:\:\frac{{dt}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}} \\ $$$$=\:\int\:\:\:\:\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {t}}{\mathrm{2}+{tan}^{\mathrm{2}} {t}}\:{dt}\:=_{{tant}\:={u}} \:\:\:\int\:\:\:\:\:\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\mathrm{2}+{u}^{\mathrm{2}} }\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\:\:\frac{{du}}{\mathrm{2}+{u}^{\mathrm{2}} }\:\:=_{{u}=\sqrt{\mathrm{2}}\alpha} \:\:\int\:\:\:\:\:\frac{\sqrt{\mathrm{2}}{d}\alpha}{\mathrm{2}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left(\frac{{u}}{\sqrt{\mathrm{2}}}\right)+{c} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left(\frac{{tant}}{\sqrt{\mathrm{2}}}\right)+{c} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left(\frac{{tan}\left(\mathrm{2}{x}\right)}{\sqrt{\mathrm{2}}}\right)\:+{c}\:\Rightarrow \\ $$$${I}\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{arctan}\left(\frac{{tan}\left(\mathrm{2}{x}\right)}{\sqrt{\mathrm{2}}}\right)\:+{c}\:. \\ $$

Answered by ajfour last updated on 03/Jun/18

∫((1+t^2 )/(1+t^4 )) dt if t=tan x =∫((1+(1/t^2 ))/((t−(1/t))^2 +((√2))^2 )) dt =(1/(√2))tan^(−1) (((tan x−cot x)/(√2)))+c =(1/(√2))tan^(−1) (−(√2)cot 2x)+c .

$$\int\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}\:\:\:\:\:{if}\:\:{t}=\mathrm{tan}\:{x} \\ $$$$=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:{dt} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:{x}−\mathrm{cot}\:{x}}{\sqrt{\mathrm{2}}}\right)+{c}\: \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(−\sqrt{\mathrm{2}}\mathrm{cot}\:\mathrm{2}{x}\right)+{c}\:. \\ $$

Commented by rahul 19 last updated on 03/Jun/18

Thanks sir ����

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