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Question Number 136969 by liberty last updated on 28/Mar/21

$${\int }_0^{\pi /2}\frac{(1+\sec {}^2\mathrm{t})\sqrt{\sec \mathrm{t}}}{{(1+\sec \mathrm{t})}^2-2}\mathrm{dt}=?$$

Answered by EDWIN88 last updated on 28/Mar/21

$$\mathrm{E}={\int }_0^{\pi /2}\frac{(1+{\sec }^2\mathrm{t})\sqrt{\sec \mathrm{t}}}{{(1+\sec \mathrm{t})}^2-2}\mathrm{dt}$$$$\mathrm{E}={\int }_0^{\pi /2}\frac{1+\cos {}^2\mathrm{t}}{\sqrt{\cos \mathrm{t}}(1+2\cos \mathrm{t}-\cos {}^2\mathrm{t})}\mathrm{dt}$$$$\mathrm{making}\mathrm{change}\mathrm{of}\mathrm{variable}\mathrm{u}=\sqrt{\cos \mathrm{t}}$$$$\mathrm{E}={\int }_1^0\frac{{\mathrm{u}}^4+1}{\mathrm{u}(1+{2\mathrm{u}}^2-{\mathrm{u}}^4)}.(-\frac{2\mathrm{u}}{\sqrt{1-{\mathrm{u}}^4}})\mathrm{du}$$$$\mathrm{E}=2{\int }_0^1\frac{{\mathrm{u}}^4+1}{(1+{2\mathrm{u}}^2-{\mathrm{u}}^4)\sqrt{1-{\mathrm{u}}^4}}\mathrm{du}$$$$\mathrm{E}=2{\int }_0^1\frac{1+{\mathrm{u}}^{-4}}{({\mathrm{u}}^{-2}+2-{\mathrm{u}}^2)\sqrt{{\mathrm{u}}^{-2}-{\mathrm{u}}^2}}\mathrm{du}$$$$\mathrm{set}\sqrt{{\mathrm{u}}^{-2}-{\mathrm{u}}^2}=\mathrm{v}$$$$\mathrm{E}=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dv}}{{\mathrm{v}}^2+2}=\sqrt{2}{\left[\arctan \left(\frac{\mathrm{v}}{\sqrt{2}}\right)\right]}_0^{\mathrm{\infty }}$$$$\mathrm{E}=\frac{\pi }{\sqrt{2}}.$$

Commented by liberty last updated on 28/Mar/21

$$\mathrm{great}$$

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