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Question Number 141507 by I want to learn more last updated on 19/May/21

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

Answered by mathmax by abdo last updated on 19/May/21

$$\mathrm{\Psi }=\int \frac{\mathrm{dx}}{{\cos }^4\mathrm{x}+{\sin }^4\mathrm{x}}\Rightarrow \mathrm{\Psi }=\int \frac{\mathrm{dx}}{{({\cos }^2\mathrm{x}+{\sin }^2\mathrm{x})}^2-{2\cos }^2\mathrm{x}.{\sin }^2\mathrm{x}}$$$$=\int \frac{\mathrm{dx}}{1-2{\left(\frac{\sin \left(2\mathrm{x}\right)}{2}\right)}^2}=\int \frac{\mathrm{dx}}{1-\frac{1}{2}{\sin }^2\left(2\mathrm{x}\right)}=\int \frac{\mathrm{dx}}{1-\frac{1-\cos \left(4\mathrm{x}\right)}{4}}$$$$=4\int \frac{\mathrm{dx}}{3+\cos \left(4\mathrm{x}\right)}{=}_{4\mathrm{x}=\mathrm{t}}\int \frac{\mathrm{dt}}{3+\mathrm{cost}}{=}_{\tan \left(\frac{\mathrm{t}}{2}\right)=\mathrm{y}}\int \frac{2\mathrm{dy}}{(1+{\mathrm{y}}^2)(3+\frac{1-{\mathrm{y}}^2}{1+{\mathrm{y}}^2})}$$$$=\int \frac{2\mathrm{dy}}{{3\mathrm{y}}^2+3+1-{\mathrm{y}}^2}=\int \frac{2\mathrm{dy}}{{2\mathrm{y}}^2+4}=\int \frac{\mathrm{dy}}{{\mathrm{y}}^2+2}{=}_{\mathrm{y}=\sqrt{2}\mathrm{z}}\int \frac{\sqrt{2}\mathrm{dz}}{2(1+{\mathrm{z}}^2)}$$$$=\frac{1}{\sqrt{2}}\arctan \left(\mathrm{z}\right)+\mathrm{C}=\frac{1}{\sqrt{2}}\arctan \left(\frac{\mathrm{y}}{\sqrt{2}}\right)+\mathrm{C}\Rightarrow$$$$\mathrm{\Psi }=\frac{1}{\sqrt{2}}\arctan \left(\frac{1}{\sqrt{2}}\tan \left(2\mathrm{x}\right)\right)+\mathrm{C}$$

Commented by I want to learn more last updated on 19/May/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}$$

Answered by peter frank last updated on 19/May/21

Answered by peter frank last updated on 19/May/21

Commented by I want to learn more last updated on 19/May/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by I want to learn more last updated on 19/May/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by Rozix last updated on 20/May/21

$$Me{l}^{\prime }mgetting\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }+\mathit{C}as$$$$myanswer.{Where}^{\prime }stheproblemplease?$$

Commented by mathmax by abdo last updated on 20/May/21

$$\mathrm{not}\mathrm{correct}\mathrm{chow}\mathrm{your}\mathrm{work}\mathrm{sir}$$

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