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Question Number 88929 by mathmax by abdo last updated on 13/Apr/20

$$cakculate{\int }_0^{\mathrm{\infty }}\frac{arctan\left(ch\left(x\right)\right)}{4+x^2}dx$$

Commented by mathmax by abdo last updated on 14/Apr/20

$$residusmethodI={\int }_0^{\mathrm{\infty }}\frac{arctan\left(chx\right)}{x^2+4}dx\Rightarrow$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(chx\right)}{x^2+4}dxletf\left(z\right)=\frac{arctan\left(chz\right)}{z^2+4}\Rightarrow$$$$f\left(z\right)=\frac{arctan\left(chz\right)}{(z-2i)(z+2i)}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=2i\pi Res(f,2i)$$$$=2i\pi \times \frac{arctan\left(ch\left(2i\right)\right)}{4i}=\frac{\pi }{2}arctan\left(ch\left(2i\right)\right)$$$$butch\left(2i\right)=\frac{e^{2i}+e^{-2i}}{2}=cos\left(2\right)\Rightarrow 2I=\frac{\pi }{2}arctan\left(cos2\right)\Rightarrow$$$$I=\frac{\pi }{4}arctan\left(cos2\right)$$$$$$

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