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Question Number 138329 by JulioCesar last updated on 12/Apr/21

Answered by Ñï= last updated on 12/Apr/21

∫(dx/(sin^4 x+cos^4 x))=∫(dx/(1−2sin^2 xcos^2 x))=∫(dx/(1−(1/2)sin^2 2x)) =∫((csc^2 2xd)/(csc^2 2x−(1/2)))=−∫((d(cot 2x))/(2cot^2 2x+1))=−tan^(−1) ((√2)cot 2x)+C

$$\int\frac{{dx}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}=\int\frac{{dx}}{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}}=\int\frac{{dx}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}} \\ $$$$=\int\frac{{csc}^{\mathrm{2}} \mathrm{2}{xd}}{{csc}^{\mathrm{2}} \mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}}=−\int\frac{{d}\left({cot}\:\mathrm{2}{x}\right)}{\mathrm{2}{cot}^{\mathrm{2}} \mathrm{2}{x}+\mathrm{1}}=−\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\mathrm{cot}\:\mathrm{2}{x}\right)+{C} \\ $$

Commented by JulioCesar last updated on 12/Apr/21

thank sir!

$${thank}\:{sir}! \\ $$

Answered by Ñï= last updated on 12/Apr/21

∫(dx/(sin^4 x+cos^4 x))=∫((tan^2 x+1)/(tan^4 x+1))d(tan x) =∫((u^2 +1)/(u^4 +1))du=∫((1+(1/u^2 ))/(u^2 +(1/u^2 )))du=∫((d(u−(1/u)))/((u−(1/u))^2 +2)) =(1/( (√2)))tan^(−1) ((u−(1/u))/( (√2)))+C=(1/( (√2)))tan^(−1) ((tan x−cot x)/( (√2)))+C

$$\int\frac{{dx}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}=\int\frac{\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{1}}{\mathrm{tan}\:^{\mathrm{4}} {x}+\mathrm{1}}{d}\left(\mathrm{tan}\:{x}\right) \\ $$$$=\int\frac{{u}^{\mathrm{2}} +\mathrm{1}}{{u}^{\mathrm{4}} +\mathrm{1}}{du}=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{du}=\int\frac{{d}\left({u}−\frac{\mathrm{1}}{{u}}\right)}{\left({u}−\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{{u}−\frac{\mathrm{1}}{{u}}}{\:\sqrt{\mathrm{2}}}+{C}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:{x}−\mathrm{cot}\:{x}}{\:\sqrt{\mathrm{2}}}+{C} \\ $$

Answered by MJS_new last updated on 12/Apr/21

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

$$\int\frac{{dx}}{\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{4}} \:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\mathrm{2}{x}\:\rightarrow\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}\:{dt}\right] \\ $$$$=\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}{t}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\mathrm{tan}\:\mathrm{2}{x}}{\:\mathrm{2}}\:+{C} \\ $$

Commented by JulioCesar last updated on 12/Apr/21

Thank sir!

$${Thank}\:{sir}! \\ $$

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