Question Number 138329 by JulioCesar last updated on 12/Apr/21
Answered by Ñï= last updated on 12/Apr/21
$$\int\frac{{dx}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}=\int\frac{{dx}}{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}}=\int\frac{{dx}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}} \\ $$$$=\int\frac{{csc}^{\mathrm{2}} \mathrm{2}{xd}}{{csc}^{\mathrm{2}} \mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}}=−\int\frac{{d}\left({cot}\:\mathrm{2}{x}\right)}{\mathrm{2}{cot}^{\mathrm{2}} \mathrm{2}{x}+\mathrm{1}}=−\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\mathrm{cot}\:\mathrm{2}{x}\right)+{C} \\ $$
Commented by JulioCesar last updated on 12/Apr/21
$${thank}\:{sir}! \\ $$
Answered by Ñï= last updated on 12/Apr/21
$$\int\frac{{dx}}{\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}}=\int\frac{\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{1}}{\mathrm{tan}\:^{\mathrm{4}} {x}+\mathrm{1}}{d}\left(\mathrm{tan}\:{x}\right) \\ $$$$=\int\frac{{u}^{\mathrm{2}} +\mathrm{1}}{{u}^{\mathrm{4}} +\mathrm{1}}{du}=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{du}=\int\frac{{d}\left({u}−\frac{\mathrm{1}}{{u}}\right)}{\left({u}−\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{{u}−\frac{\mathrm{1}}{{u}}}{\:\sqrt{\mathrm{2}}}+{C}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:{x}−\mathrm{cot}\:{x}}{\:\sqrt{\mathrm{2}}}+{C} \\ $$
Answered by MJS_new last updated on 12/Apr/21
$$\int\frac{{dx}}{\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{4}} \:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\mathrm{2}{x}\:\rightarrow\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}\:{dt}\right] \\ $$$$=\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}{t}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\mathrm{tan}\:\mathrm{2}{x}}{\:\mathrm{2}}\:+{C} \\ $$
Commented by JulioCesar last updated on 12/Apr/21
$${Thank}\:{sir}! \\ $$