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Question Number 109909 by mathdave last updated on 26/Aug/20

Commented by mnjuly1970 last updated on 26/Aug/20

$$niceproblembutalittlebit$$$$difficult.★$$

Commented by mathdave last updated on 26/Aug/20

$$itverysimpleproblem$$

Commented by mohammad17 last updated on 26/Aug/20

$$siriwantthesolutionthisproblem$$

Commented by mathmax by abdo last updated on 26/Aug/20

$$\mathrm{if}\mathrm{you}\mathrm{have}\mathrm{seen}\mathrm{the}\mathrm{answer}\mathrm{in}\mathrm{a}\mathrm{book}...$$

Commented by mathdave last updated on 26/Aug/20

$$whatdidyoureallymeantbythatstatement$$

Commented by mathmax by abdo last updated on 26/Aug/20

$$\mathrm{i}\mathrm{mean}\mathrm{that}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{not}\mathrm{simple}...!$$

Answered by mnjuly1970 last updated on 26/Aug/20

Answered by mathmax by abdo last updated on 26/Aug/20

$$\mathrm{I}={\int }_0^{\pi }\cos (\mathrm{tanx}-\frac{1}{\mathrm{tanx}})\mathrm{dx}\Rightarrow$$$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\cos (\mathrm{tanx}-\frac{1}{\mathrm{tanx}})\mathrm{dx}+{\int }_{\frac{\pi }{2}}^{\pi }\cos (\mathrm{tanx}-\frac{1}{\mathrm{tanx}})\mathrm{dx}\mathrm{but}$$$${\int }_{\frac{\pi }{2}}^{\pi }\cos (\mathrm{tanx}-\frac{1}{\mathrm{tanx}})\mathrm{dx}{=}_{\mathrm{x}=\frac{\pi }{2}+\mathrm{t}}{\int }_0^{\frac{\pi }{2}}\cos (-\frac{1}{\mathrm{tant}}+\mathrm{tant})\mathrm{dt}$$$$={\int }_0^{\frac{\pi }{2}}\cos (\mathrm{tant}-\frac{1}{\mathrm{tant}})\mathrm{dt}\Rightarrow \mathrm{I}=2{\int }_0^{\frac{\pi }{2}}\cos (\mathrm{tanx}-\frac{1}{\mathrm{tanx}})\mathrm{dx}$$$${=}_{\mathrm{tanx}=\mathrm{u}}2{\int }_0^{\mathrm{\infty }}\cos (\mathrm{u}-\frac{1}{\mathrm{u}})\frac{\mathrm{du}}{1+{\mathrm{u}}^2}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\cos (\mathrm{u}-\frac{1}{\mathrm{u}})}{{\mathrm{u}}^2+1}\mathrm{du}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos (\mathrm{u}-\frac{1}{\mathrm{u}})}{{\mathrm{u}}^2+1}\mathrm{du}=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{i}(\mathrm{u}-\frac{1}{\mathrm{u}})}}{{\mathrm{u}}^2+1}\mathrm{du}\right)$$$$\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{i}(\mathrm{z}-\frac{1}{\mathrm{z}})}}{{\mathrm{z}}^2+1}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi -\mathrm{z})\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})$$$$=2\mathrm{i}\pi \times \frac{{\mathrm{e}}^{\mathrm{i}(\mathrm{i}-\frac{1}{\mathrm{i}})}}{2\mathrm{i}}=\pi {\mathrm{e}}^{-1-1}=\pi {\mathrm{e}}^{-2}\Rightarrow ★{\int }_0^{\pi }\cos (\mathrm{tanx}-\mathrm{cotanx})\mathrm{dx}=\frac{\pi }{{\mathrm{e}}^2}★$$$$$$

Commented by mathmax by abdo last updated on 26/Aug/20

$$\mathrm{the}\mathrm{Q}\mathrm{here}\mathrm{is}\mathrm{calculate}{\int }_0^{\pi }\cos (\mathrm{tanx}-\mathrm{cotanx})\mathrm{dx}$$

Commented by mnjuly1970 last updated on 26/Aug/20

$$veryexcellent...$$

Commented by mathmax by abdo last updated on 26/Aug/20

$$\mathrm{thanks}\mathrm{sir}$$

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