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Question Number 73338 by mathmax by abdo last updated on 10/Nov/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{arctan\left(2cosx\right)}{3+x^2}dx$$

Commented by mathmax by abdo last updated on 11/Nov/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{arctan\left(2cosx\right)}{x^2+3}dx\Rightarrow I{=}_{x=\sqrt{3}t}{\int }_0^{\mathrm{\infty }}\frac{arctan\left(2cos\left(\sqrt{3}t\right)\right)}{3(t^2+1)}\sqrt{3}dt$$$$=\frac{1}{\sqrt{3}}{\int }_0^{\mathrm{\infty }}\frac{arctan\left(2cos\left(\sqrt{3}t\right)\right)}{t^2+1}dt\Rightarrow 2\sqrt{3}I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan(2cos\left(\sqrt{3}t\right)}{t^2+1}dt$$$$letw\left(z\right)=\frac{arctan\left(2cos\left(\sqrt{3}z\right)\right)}{z^2+1}=\frac{arctan\left(2cos\left(\sqrt{3}z\right)\right)}{(z-i)(z+i)}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}w\left(z\right)dz=2i\pi Res(w,i)$$$$Res(w,i)=\frac{arctan\left(2cos\left(i\sqrt{3}\right)\right)}{2i}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}w\left(z\right)dz=2i\pi \times \frac{arctan\left(2cos\left(i\sqrt{3}\right)\right)}{2i}$$$$=\pi arctan\left(2cos\left(i\sqrt{3}\right)\right)but$$$$cos\left(i\sqrt{3}\right)=ch\left(\sqrt{3}\right)=\frac{e^{\sqrt{3}}+e^{-\sqrt{3}}}{2}\Rightarrow arctan\left(2cos\left(i\sqrt{3}\right)\right)$$$$=arctan(e^{\sqrt{3}}+e^{-\sqrt{3}})\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}w\left(\right)dz=\pi arctan(e^{\sqrt{3}}+e^{-\sqrt{3}})\Rightarrow$$$$I=\frac{\pi }{2\sqrt{3}}arctan(e^{\sqrt{3}}+e^{-\sqrt{3}}).$$

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