Question Number 152730 by puissant last updated on 31/Aug/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{{tanx}}{\:\sqrt{\mathrm{2}{cosx}−\mathrm{1}}}{dx} \\ $$
Answered by Olaf_Thorendsen last updated on 31/Aug/21
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \frac{\mathrm{tan}{x}}{\:\sqrt{\mathrm{2cos}{x}−\mathrm{1}}} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{cos}{x}\:: \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\frac{−{du}}{{u}}}{\:\sqrt{\mathrm{2}{u}−\mathrm{1}}} \\ $$$$\mathrm{I}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{du}}{\:{u}\sqrt{\mathrm{2}{u}−\mathrm{1}}} \\ $$$$\mathrm{I}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\frac{\mathrm{2}{du}}{\mathrm{2}\sqrt{\mathrm{2}{u}−\mathrm{1}}}}{\:{u}} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\frac{\mathrm{2}{du}}{\mathrm{2}\sqrt{\mathrm{2}{u}−\mathrm{1}}}}{\:\mathrm{1}+\left(\sqrt{\mathrm{2}{u}−\mathrm{1}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\:=\:\mathrm{2}\left[\mathrm{arctan}\left(\sqrt{\mathrm{2}{u}−\mathrm{1}}\right)\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\left(\frac{\pi}{\mathrm{4}}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$
Commented by Olaf_Thorendsen last updated on 31/Aug/21