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Question Number 133036 by liberty last updated on 18/Feb/21

$${\underset{0}{\int }}^{\frac{\pi }{2}}\frac{dx}{1+\tan {}^{2014}\left(x\right)}=\frac{\pi {\mathrm{e}}^{\mathrm{q}}}{\mathrm{p}}$$$$\mathrm{Find}2\mathrm{p}-\mathrm{q}.$$

Answered by liberty last updated on 18/Feb/21

$$\mathrm{I}={\underset{0}{\int }}^{\frac{\pi }{2}}\frac{\mathrm{dx}}{1+\tan {}^{2014}\left(\mathrm{x}\right)}={\underset{\frac{\pi }{2}}{\int }}^0\frac{\tan {}^{2014}\left(\mathrm{x}\right)}{1+\tan {}^{2014}\left(\mathrm{x}\right)}(-\mathrm{dx})$$$$\mathrm{I}={\underset{0}{\int }}^{\frac{\pi }{2}}\frac{\tan {}^{2014}\left(\mathrm{x}\right)}{1+\tan {}^{2014}\left(\mathrm{x}\right)}\mathrm{dx}$$$$\mathrm{we}\mathrm{get}2\mathrm{I}={\underset{0}{\int }}^{\frac{\pi }{2}}\frac{1+\tan {}^{2014}\left(\mathrm{x}\right)}{1+\tan {}^{2014}\left(\mathrm{x}\right)}\mathrm{dx}$$$$\mathrm{I}=\frac{1}{2}.\frac{\pi }{2}=\frac{\pi }{4}=\frac{\pi {\mathrm{e}}^{\mathrm{q}}}{\mathrm{p}}$$$$\to \{\begin{array}{l}\mathrm{q}=0\\ \mathrm{p}=4\end{array}\Rightarrow 2\mathrm{p}-\mathrm{q}=8$$

Answered by mathmax by abdo last updated on 20/Feb/21

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{1+{\tan }^{2014}\mathrm{x}}\Rightarrow \mathrm{I}{=}_{\mathrm{tanx}=\mathrm{z}}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dz}}{(1+{\mathrm{z}}^2)(1+{\mathrm{z}}^{2014})}$$$$\mathrm{I}{=}_{\mathrm{z}=\frac{1}{\mathrm{t}}}-{\int }_0^{\mathrm{\infty }}\frac{-\mathrm{dt}}{{\mathrm{t}}^2(1+\frac{1}{{\mathrm{t}}^2})(1+\frac{1}{{\mathrm{t}}^{2014}})}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{2014}}{(1+{\mathrm{t}}^2)(1+{\mathrm{t}}^{2014})}\Rightarrow$$$$2\mathrm{I}={\int }_0^{\mathrm{\infty }}(\frac{1}{(1+{\mathrm{z}}^2)(1+{\mathrm{z}}^{2014})}+\frac{{\mathrm{z}}^{2014}}{(1+{\mathrm{z}}^2)(1+{\mathrm{z}}^{2014})})\mathrm{dz}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dz}}{1+{\mathrm{z}}^2}$$$$=\frac{\pi }{2}\Rightarrow \mathrm{I}=\frac{\pi }{4}=\frac{\pi }{\mathrm{p}}{\mathrm{e}}^{\mathrm{q}}\Rightarrow \mathrm{p}=4\mathrm{and}\mathrm{q}=0\Rightarrow 2\mathrm{p}-\mathrm{q}=8$$

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