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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

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

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

$$thankyou$$

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

$$youarewelcome.$$

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

$${\sin }^2\theta +{\cos }^2\theta =1$$$${({\sin }^2\theta +{\cos }^2\theta )}^2=1^2$$$${\sin }^4\theta +{\cos }^4\theta +2{\sin }^2\theta {\cos }^2\theta =1$$$${\sin }^4\theta +{\cos }^4\theta +\frac{{\sin }^{2}2\theta }{2}=1$$$${\sin }^4\theta +{\cos }^4\theta =\frac{2-{\sin }^{2}2\theta }{2}$$$${\sin }^4\theta +{\cos }^4\theta =\frac{(\sqrt{2}-\sin 2\theta )(\sqrt{2}+\sin 2\theta )}{2}$$$$\int \frac{d\theta }{{\sin }^4\theta +{\cos }^4\theta }$$$$=\int \frac{2d\theta }{(\sqrt{2}-\sin 2\theta )(\sqrt{2}+\sin 2\theta )}$$$$=\int \frac{du}{(\sqrt{2}-\sin u)(\sqrt{2}+\sin u)}withu=2\theta$$$$=\frac{1}{2\sqrt{2}}\int [\frac{1}{\sqrt{2}-\sin u}+\frac{1}{\sqrt{2}+\sin u}]du$$$$=\frac{1}{2\sqrt{2}}[2{\tan }^{-1}(\sqrt{2}\tan \frac{u}{2}-1)+2{\tan }^{-1}(\sqrt{2}\tan \frac{u}{2}+1)]+C$$$$=\frac{1}{\sqrt{2}}[{\tan }^{-1}(\sqrt{2}\tan \theta -1)+{\tan }^{-1}(\sqrt{2}\tan \theta +1)]+C$$

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

$$thankyou$$

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

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

Answered by vishalbhardwaj last updated on 24/Dec/19

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

Commented by malwaan last updated on 15/Dec/19

$$\mathit{f}\mathit{a}\mathit{n}\mathit{t}\mathit{a}\mathit{s}\mathit{t}\mathit{i}\mathit{c}$$$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{x}4\mathit{a}\mathit{l}\mathit{l}$$

Answered by MJS last updated on 15/Dec/19

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

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

$$thankyou$$

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

$$sorrysiripressing$$$$flagoptionbymistake$$

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