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Question Number 31515 by abdo imad last updated on 09/Mar/18

$$calculate{\int }_0^1\frac{dx}{chx}.$$

Commented by abdo imad last updated on 10/Mar/18

$$I={\int }_0^1\frac{dx}{\frac{e^x+e^{-x}}{2}}=2{\int }_0^1\frac{dx}{e^x+e^{-x}}thech.e^x=tgive$$$$I=2{\int }_1^e\frac{1}{t+\frac{1}{t}}\frac{dt}{t}=2{\int }_1^e\frac{dt}{t^2+1}=2{\left[arctant\right]}_1^e=2arctane-\frac{\pi }{2}.$$

Answered by sma3l2996 last updated on 10/Mar/18

$$I={\int }_0^1\frac{dx}{cosh\left(x\right)}$$$$t=tanh\left(x/2\right)\Rightarrow 2dt=(1-{\left(tanh\left(x/2\right)\right)}^2)dx$$$$cosh\left(x\right)=2{cosh}^2\left(x/2\right)-1=\frac{2}{1-{tanh}^2\left(x/2\right)}-1$$$$cosh\left(x\right)=\frac{1+{tanh}^2\left(x/2\right)}{1-{tanh}^2\left(x/2\right)}=\frac{1+t^2}{1-t^2}$$$$I={\int }_0^{tanh\left(1/2\right)}\frac{1-t^2}{1+t^2}\times \left(\frac{2dt}{1-t^2}\right)=2{\int }_0^{tanh\left(1/2\right)}\frac{dt}{1+t^2}$$$$I=2{\left[{tan}^{-1}\left(t\right)\right]}_0^{tanh\left(1/2\right)}$$$$I=2{tan}^{-1}\left(tanh\left(1/2\right)\right)$$

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