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Question Number 168366 by cortano1 last updated on 09/Apr/22

Answered by qaz last updated on 09/Apr/22

$${\int }_{-\pi /2}^{\pi /2}\frac{\cos \mathrm{x}}{1-\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx}={\int }_{-\pi /2}^{\pi /2}\frac{\cos \mathrm{x}}{1+\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx}$$$$=\frac{1}{2}{\int }_{-\pi /2}^{\pi /2}\frac{2(1+\sqrt{{\mathrm{x}}^2+1})\cos \mathrm{x}}{{(1+\sqrt{{\mathrm{x}}^2+1})}^2-{\mathrm{x}}^2}\mathrm{dx}=\frac{1}{2}{\int }_{-\pi /2}^{\pi /2}\frac{(1+\sqrt{{\mathrm{x}}^2+1})\cos \mathrm{x}}{1+\sqrt{{\mathrm{x}}^2+1}}\mathrm{dx}$$$$={\int }_0^{\pi /2}\cos \mathrm{xdx}=1$$

Commented by peter frank last updated on 09/Apr/22

$$\mathrm{thanks}$$

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