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Question Number 158590 by mnjuly1970 last updated on 06/Nov/21

$$$$$$provethat$$$$1.\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{sin(x+tan\left(x\right))}{sin\left(x\right)}dx=\frac{\pi }{2}$$$$2.\mathrm{J}={\int }_0^{\frac{\pi }{2}}\frac{sin(x-tan\left(x\right))}{sin\left(x\right)}dx=(\frac{1}{e}-\frac{1}{2})\pi$$$$$$

Answered by qaz last updated on 06/Nov/21

$$\mathrm{I}={\int }_0^{\pi /2}\frac{\sin \mathrm{x}\cdot \cos \left(\tan \mathrm{x}\right)+\cos \mathrm{x}\cdot \sin \left(\tan \mathrm{x}\right)}{\sin \mathrm{x}}\mathrm{dx}$$$$={\int }_0^{\pi /2}\cos \left(\tan \mathrm{x}\right)+\frac{\sin \left(\tan \mathrm{x}\right)}{\tan \mathrm{x}}\mathrm{dx}............\tan \mathrm{x}\to \mathrm{x}$$$$={\int }_0^{\mathrm{\infty }}(\cos \mathrm{x}+\frac{\sin \mathrm{x}}{\mathrm{x}})\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$$$={\int }_0^{\mathrm{\infty }}\frac{\cos \mathrm{x}}{1+{\mathrm{x}}^2}+(\frac{1}{\mathrm{x}}-\frac{\mathrm{x}}{1+{\mathrm{x}}^2})\sin \mathrm{xdx}$$$$={\int }_0^{\mathrm{\infty }}\mathcal{L}\left\{\cos \mathrm{x}\right\}\cdot {\mathcal{L}}^{-1}\left\{\frac{1}{1+{\mathrm{x}}^2}\right\}+(\frac{1}{\mathrm{x}}-\frac{\mathrm{x}}{1+{\mathrm{x}}^2})\sin \mathrm{xdx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{x}}{1+{\mathrm{x}}^2}\cdot \sin \mathrm{x}+(\frac{1}{\mathrm{x}}-\frac{\mathrm{x}}{1+{\mathrm{x}}^2})\cdot \sin \mathrm{xdx}$$$$=\frac{\pi }{2}$$$$---------------$$$$\mathrm{J}={\int }_0^{\pi /2}\frac{\sin (\mathrm{x}-\tan \mathrm{x})}{\sin \mathrm{x}}\mathrm{dx}$$$$={\int }_0^{\pi /2}\frac{\sin \mathrm{x}\cdot \cos \left(\tan \mathrm{x}\right)-\cos \mathrm{x}\cdot \sin \left(\tan \mathrm{x}\right)}{\sin \mathrm{x}}\mathrm{dx}$$$$={\int }_0^{\pi /2}\cos \left(\tan \mathrm{x}\right)-\frac{\sin \left(\tan \mathrm{x}\right)}{\tan \mathrm{x}}\mathrm{dx}.........\tan \mathrm{x}\to \mathrm{x}$$$$={\int }_0^{\mathrm{\infty }}(\cos \mathrm{x}-\frac{\sin \mathrm{x}}{\mathrm{x}})\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$$$={\int }_0^{\mathrm{\infty }}(\frac{2\cos \mathrm{x}}{1+{\mathrm{x}}^2}-\frac{\sin \mathrm{x}}{\mathrm{x}})\mathrm{dx}$$$$=\frac{\pi }{\mathrm{e}}-\frac{\pi }{2}$$$$-----------$$$${\int }_0^{\mathrm{\infty }}\frac{\sin \mathrm{x}}{\mathrm{x}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\mathcal{L}\left\{\sin \mathrm{x}\right\}\cdot {\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{x}}\right\}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{1}{1+{\mathrm{x}}^2}\cdot 1\mathrm{dx}=\frac{\pi }{2}$$$$------------$$$${\int }_0^{\mathrm{\infty }}\frac{\mathcal{L}\left\{\cos \mathrm{tx}\right\}}{1+{\mathrm{x}}^2}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{s}}{({\mathrm{s}}^2+{\mathrm{x}}^2)(1+{\mathrm{x}}^2)}\mathrm{dx}=\frac{\pi }{2(1+\mathrm{s})}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\cos \mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}={\mathcal{L}}^{-1}{\left\{\frac{\pi }{2(1+\mathrm{s})}\right\}}_{\mathrm{t}=1}=\frac{\pi }{2}{\mathrm{e}}^{-1}$$

Commented by puissant last updated on 06/Nov/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{cosx}{x^2+1}dx=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cosx}{x^2+1}dx$$$$=\frac{1}{2}Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{ix}}{x^2+1}dx\right)=\frac{1}{2}(2i\pi \bullet res(\frac{e^{ix}}{x^2+1},+i))$$$$\mathrm{\Omega }=i\pi \bullet \underset{x\to +i}{\lim }\left(\frac{e^{i\pi }}{x+i}\right)=i\pi \times \frac{e^{-1}}{2i}=\frac{\pi }{2e}..$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{cosx}{x^2+1}dx=\frac{\pi }{2e}..$$$$............\mathcal{L}epuissant............$$

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