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Question-201027




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

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