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Question Number 127885 by psyche last updated on 02/Jan/21

$$\int \left(\frac{\sin ({2\tan }^{-1}\left(x\right)+x)}{x}\right)thelimit[0,\mathrm{\infty })$$

Answered by Lordose last updated on 02/Jan/21

$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\sin ({2\tan }^{-1}\left(\mathrm{x}\right)+\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$\sin ({2\tan }^{-1}\left(\mathrm{x}\right)+\mathrm{x})=\sin \left({2\tan }^{-1}\left(\mathrm{x}\right)\right)\cos \left(\mathrm{x}\right)+\cos \left({2\tan }^{-1}\left(\mathrm{x}\right)\right)\sin \left(\mathrm{x}\right)$$$$\mathrm{N}.\mathrm{B}::\left(\begin{array}{c}\sin \left({2\tan }^{-1}\left(\mathrm{x}\right)\right)=\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\\ \cos \left({2\tan }^{-1}\left(\mathrm{x}\right)\right)=\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\end{array}\right)$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\frac{1}{1+{\mathrm{x}}^2}(2\mathrm{xcos}\left(\mathrm{x}\right)+(1-{\mathrm{x}}^2)\sin \left(\mathrm{x}\right))}{\mathrm{x}}$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{2\cos \left(\mathrm{x}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}+{\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{x}\right)}{\mathrm{x}(1+{\mathrm{x}}^2)}\mathrm{dx}-{\int }_0^{\mathrm{\infty }}\frac{\mathrm{xsin}\left(\mathrm{x}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{\Omega }=2\cdot \frac{\pi }{2\mathrm{e}}+{\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{x}\right)}{\mathrm{x}(1+{\mathrm{x}}^2)}\mathrm{dx}-{\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}+{\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{x}\right)}{\mathrm{x}(1+{\mathrm{x}}^2)}\mathrm{dx}$$$$\mathrm{\Omega }=\frac{\pi }{\mathrm{e}}-\frac{\pi }{2}+2\mathrm{\Phi }$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{x}\right)}{\mathrm{x}(1+{\mathrm{x}}^2)}\mathrm{dx}=\frac{\pi }{2}(1-{\mathrm{e}}^{-1})$$$$\mathrm{\Omega }=\frac{\pi }{\mathrm{e}}-\frac{\pi }{2}+\pi -\frac{\pi }{\mathrm{e}}=\frac{\pi }{2}$$$$★\mathbf{L}\mathit{\varphi }\mathbf{rD}\mathit{\varnothing }\mathbf{sE}$$

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