Question Number 201027 by sonukgindia last updated on 28/Nov/23
Answered by Atomist last updated on 28/Nov/23
$${sechx}=\frac{\mathrm{1}}{{coshx}}\: \\ $$$${coshx}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$\int{sechxdx}=\int\frac{\mathrm{2}}{{e}^{{x}} +{e}^{−{x}} }{dx}= \\ $$$$=\int\frac{\mathrm{2}}{{e}^{{x}} +{e}^{−{x}} }×\frac{{e}^{{x}} }{{e}^{{x}} }{dx}= \\ $$$$\int\frac{\mathrm{2}{e}^{{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{1}}{dx}= \\ $$$${u}={e}^{{x}} \\ $$$$\mathrm{2}{dv}=\mathrm{2}{e}^{{x}} {dx} \\ $$$$=\int\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}=\mathrm{2}\int\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }= \\ $$$$\int\frac{{dx}}{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}}{arctg}\frac{{x}}{{a}}+{C} \\ $$$$=\mathrm{2}{arctg}\left({e}^{{x}} \right)+{C} \\ $$$$\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{2}} \\ $$$${Answer}.\:\frac{\pi}{\mathrm{2}} \\ $$