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Question Number 183706 by cortano1 last updated on 29/Dec/22

$$A=\int \frac{1+\cos {}^{2}x}{1+\sin {}^{4}x}dx$$

Answered by Ar Brandon last updated on 29/Dec/22

$$A=\int \frac{1+{\cos }^{2}x}{1+{\sin }^{4}x}dx=\int \frac{{\sec }^{4}x+{\sec }^{2}x}{{\sec }^{4}x+{\tan }^{4}x}dx=\int \frac{({\tan }^{2}x+2){\sec }^{2}x}{{2\tan }^{4}x+{2\tan }^{2}x+1}dx$$$$=\int \frac{t^2+2}{2t^4+2t^2+1}dt=\frac{1}{2}\int \frac{\frac{1+2\sqrt{2}}{2}(t^2+\frac{1}{\sqrt{2}})+\frac{1-2\sqrt{2}}{2}(t^2-\frac{1}{\sqrt{2}})}{t^4+t^2+\frac{1}{2}}dt$$$$=\frac{1+2\sqrt{2}}{4}\int \frac{t^2+\frac{1}{\sqrt{2}}}{t^4+t^2+\frac{1}{2}}dt+\frac{1-2\sqrt{2}}{4}\int \frac{t^2-\frac{1}{\sqrt{2}}}{t^4+t^2+\frac{1}{2}}dt$$$$=\frac{1+2\sqrt{2}}{4}\int \frac{1+\frac{1}{\sqrt{2}{t}^2}}{t^2+1+\frac{1}{2t^2}}dt+\frac{1-2\sqrt{2}}{4}\int \frac{1-\frac{1}{\sqrt{2}{t}^2}}{t^2+1+\frac{1}{2t^2}}dt$$$$=\frac{1+2\sqrt{2}}{4}\int \frac{1+\frac{1}{\sqrt{2}{t}^2}}{{(t-\frac{1}{\sqrt{2}t})}^2+1+\frac{2}{\sqrt{2}}}dt+\frac{1-2\sqrt{2}}{4}\int \frac{1-\frac{1}{\sqrt{2}{t}^2}}{{(t+\frac{1}{\sqrt{2}t})}^2+1-\frac{2}{\sqrt{2}}}dt$$$$=\frac{1+2\sqrt{2}}{4}\int \frac{du}{u^2+\frac{\sqrt{2}+2}{\sqrt{2}}}+\frac{1-2\sqrt{2}}{4}\int \frac{d\mathrm{v}}{{\mathrm{v}}^2+\frac{\sqrt{2}-2}{\sqrt{2}}}$$$$=\frac{1+2\sqrt{2}}{4}\cdot \sqrt{\frac{\sqrt{2}}{\sqrt{2}+2}}\arctan \left(u\sqrt{\frac{\sqrt{2}}{\sqrt{2}+2}}\right)+\frac{1-2\sqrt{2}}{4}\cdot \sqrt{\frac{\sqrt{2}}{\sqrt{2}-2}}\arctan \left(\mathrm{v}\sqrt{\frac{\sqrt{2}}{\sqrt{2}-2}}\right)+C$$$$=\frac{1+2\sqrt{2}}{4}\sqrt{\frac{\sqrt{2}}{\sqrt{2}+2}}\arctan \left(\frac{\sqrt{2}{\tan }^{2}x-1}{\sqrt{2}\tan x}\sqrt{\frac{\sqrt{2}}{\sqrt{2}+2}}\right)+\frac{1-2\sqrt{2}}{4}\sqrt{\frac{\sqrt{2}}{\sqrt{2}-2}}\arctan \left(\frac{\sqrt{2}{\tan }^{2}x+1}{\sqrt{2}\tan x}\sqrt{\frac{\sqrt{2}}{\sqrt{2}-2}}\right)+C$$

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