Question Number 31515 by abdo imad last updated on 09/Mar/18
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{{chx}}\:. \\ $$
Commented by abdo imad last updated on 10/Mar/18
![I= ∫_0 ^1 (dx/((e^x +e^(−x) )/2))=2 ∫_0 ^1 (dx/(e^x +e^(−x) )) the ch .e^x =t give I=2 ∫_1 ^e (1/(t +(1/t))) (dt/t)= 2∫_1 ^e (dt/(t^2 +1))=2[arctant]_1 ^e =2 arctane −(π/2) .](https://www.tinkutara.com/question/Q31581.png)
$${I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\frac{{e}^{{x}} \:+{e}^{−{x}} }{\mathrm{2}}}=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{{e}^{{x}} \:+{e}^{−{x}} }\:{the}\:{ch}\:.{e}^{{x}} ={t}\:{give} \\ $$$${I}=\mathrm{2}\:\int_{\mathrm{1}} ^{{e}} \:\:\:\:\:\frac{\mathrm{1}}{{t}\:+\frac{\mathrm{1}}{{t}}}\:\frac{{dt}}{{t}}=\:\mathrm{2}\int_{\mathrm{1}} ^{{e}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2}\left[{arctant}\right]_{\mathrm{1}} ^{{e}} \:=\mathrm{2}\:{arctane}\:−\frac{\pi}{\mathrm{2}}\:. \\ $$
Answered by sma3l2996 last updated on 10/Mar/18
![I=∫_0 ^1 (dx/(cosh(x))) t=tanh(x/2)⇒2dt=(1−(tanh(x/2))^2 )dx cosh(x)=2cosh^2 (x/2)−1=(2/(1−tanh^2 (x/2)))−1 cosh(x)=((1+tanh^2 (x/2))/(1−tanh^2 (x/2)))=((1+t^2 )/(1−t^2 )) I=∫_0 ^(tanh(1/2)) ((1−t^2 )/(1+t^2 ))×(((2dt)/(1−t^2 )))=2∫_0 ^(tanh(1/2)) (dt/(1+t^2 )) I=2[tan^(−1) (t)]_0 ^(tanh(1/2)) I=2tan^(−1) (tanh(1/2))](https://www.tinkutara.com/question/Q31554.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{cosh}\left({x}\right)} \\ $$$${t}={tanh}\left({x}/\mathrm{2}\right)\Rightarrow\mathrm{2}{dt}=\left(\mathrm{1}−\left({tanh}\left({x}/\mathrm{2}\right)\right)^{\mathrm{2}} \right){dx} \\ $$$${cosh}\left({x}\right)=\mathrm{2}{cosh}^{\mathrm{2}} \left({x}/\mathrm{2}\right)−\mathrm{1}=\frac{\mathrm{2}}{\mathrm{1}−{tanh}^{\mathrm{2}} \left({x}/\mathrm{2}\right)}−\mathrm{1} \\ $$$${cosh}\left({x}\right)=\frac{\mathrm{1}+{tanh}^{\mathrm{2}} \left({x}/\mathrm{2}\right)}{\mathrm{1}−{tanh}^{\mathrm{2}} \left({x}/\mathrm{2}\right)}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$${I}=\int_{\mathrm{0}} ^{{tanh}\left(\mathrm{1}/\mathrm{2}\right)} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }×\left(\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\right)=\mathrm{2}\int_{\mathrm{0}} ^{{tanh}\left(\mathrm{1}/\mathrm{2}\right)} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${I}=\mathrm{2}\left[{tan}^{−\mathrm{1}} \left({t}\right)\right]_{\mathrm{0}} ^{{tanh}\left(\mathrm{1}/\mathrm{2}\right)} \\ $$$${I}=\mathrm{2}{tan}^{−\mathrm{1}} \left({tanh}\left(\mathrm{1}/\mathrm{2}\right)\right) \\ $$