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Question Number 201293 by sonukgindia last updated on 03/Dec/23

Answered by aleks041103 last updated on 03/Dec/23

$$I={\int }_2^{\mathrm{\infty }}\frac{8arcsec\left(x/2\right)dx}{x^3-4x}=$$$$={\int }_1^{\mathrm{\infty }}\frac{8arcsec\left(\left(2x\right)/2\right)}{{\left(2x\right)}^3-4\left(2x\right)}d\left(2x\right)=2{\int }_1^{\mathrm{\infty }}\frac{arcsec\left(x\right)}{x(x^2-1)}dx$$$$x=sec\left(t\right)\Rightarrow dx=sec\left(t\right)tan\left(t\right)dt$$$${\int }_0^{\pi /2}\frac{tsec\left(t\right)tan\left(t\right)dt}{sec\left(t\right)({sec}^2\left(t\right)-1)}=$$$$={\int }_0^{\pi /2}\frac{tcos\left(t\right)}{sin\left(t\right)}dt={\int }_0^{\pi /2}td\left(ln\left(sin\left(t\right)\right)\right)=$$$$={\left[tln\left(sin\left(t\right)\right)\right]}_0^{\pi /2}-{\int }_0^{\pi /2}ln\left(sin\left(t\right)\right)dt$$$$=-{\int }_0^{\pi /2}ln\left(sin\left(t\right)\right)dt$$$$J={\int }_0^{\pi /2}ln\left(sin\left(t\right)\right)dt={\int }_{\pi /2}^{0}ln\left(sin(\frac{\pi }{2}-s)\right)d(\frac{\pi }{2}-s)=$$$$={\int }_0^{\pi /2}ln\left(cos\left(s\right)\right)ds$$$$\Rightarrow 2J=J+J={\int }_0^{\pi /2}(ln\left(sin\left(x\right)\right)+ln\left(cos\left(x\right)\right))dx=$$$$={\int }_0^{\pi /2}ln\left(sin\left(x\right)cos\left(x\right)\right)dx=$$$$={\int }_0^{\pi /2}ln\left(\frac{sin\left(2x\right)}{2}\right)dx=$$$$={\int }_0^{\pi /2}ln\left(sin\left(2x\right)\right)dx-\frac{\pi }{2}ln\left(2\right)=$$$$=\frac{1}{2}{\int }_0^{\pi }ln\left(sin\left(x\right)\right)dx-\frac{\pi ln\left(2\right)}{2}=$$$$=\frac{1}{2}{\int }_{-\pi /2}^{\pi /2}ln\left(cos\left(x\right)\right)dx-\frac{\pi ln\left(2\right)}{2}=$$$$={\int }_0^{\pi /2}ln\left(cos\left(x\right)\right)dx-\frac{\pi ln\left(2\right)}{2}=J-\frac{\pi ln\left(2\right)}{2}$$$$\Rightarrow J=-\frac{\pi ln\left(2\right)}{2}$$$$\Rightarrow I=-2J$$$$\Rightarrow {\int }_2^{\mathrm{\infty }}\frac{8arcsec\left(x/2\right)}{x^3-4x}dx=\pi ln\left(2\right)$$

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