Question Number 134185 by mnjuly1970 last updated on 28/Feb/21
![.....advanced calculus.... prove that:: i:π=β«_0 ^( (Ο/2)) ((cos(tan(x)βx))/(cos(x)))dx=(Ο/e) ii:Ξ _(n=2) ^β e(1β(1/n^2 ))^n^2 =(Ο/(e(βe)))](https://www.tinkutara.com/question/Q134185.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:…..{advanced}\:\:\:{calculus}…. \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:{i}:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({tan}\left({x}\right)β{x}\right)}{{cos}\left({x}\right)}{dx}=\frac{\pi}{{e}} \\ $$$$\:\:\:\:{ii}:\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}{e}\left(\mathrm{1}β\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{{n}^{\mathrm{2}} } =\frac{\pi}{{e}\sqrt{{e}}} \\ $$$$\:\: \\ $$
Answered by mindispower last updated on 28/Feb/21
![β
=β«_0 ^(Ο/2) cos(tg(x))+sin(tg(x))tg(x)dx =A+B,tg(x)=tβ A=(1/2)Reβ«_(ββ) ^β (e^(it) /(1+t^2 ))dt=Re(iΟ.Res((e^(it) /(1+t^2 )),t=i) =(Ο/(2e)) B=(1/2)Imβ«_(ββ) ^β (t/(1+t^2 ))e^(it) dt =Im.iΟRes(((te^(it) )/(1+t^2 )),t=i)=(Ο/(2e)) β
=A+B=(Ο/e)](https://www.tinkutara.com/question/Q134192.png)
$$\emptyset=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left({tg}\left({x}\right)\right)+{sin}\left({tg}\left({x}\right)\right){tg}\left({x}\right){dx} \\ $$$$={A}+{B},{tg}\left({x}\right)={t}\Rightarrow \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}{Re}\int_{β\infty} ^{\infty} \frac{{e}^{{it}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}={Re}\left({i}\pi.{Res}\left(\frac{{e}^{{it}} }{\mathrm{1}+{t}^{\mathrm{2}} },{t}={i}\right)\right. \\ $$$$=\frac{\pi}{\mathrm{2}{e}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}}{Im}\int_{β\infty} ^{\infty} \frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{e}^{{it}} {dt} \\ $$$$={Im}.{i}\pi{Res}\left(\frac{{te}^{{it}} }{\mathrm{1}+{t}^{\mathrm{2}} },{t}={i}\right)=\frac{\pi}{\mathrm{2}{e}} \\ $$$$\emptyset={A}+{B}=\frac{\pi}{{e}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 28/Feb/21
![mercey mr mindispower..](https://www.tinkutara.com/question/Q134196.png)
$${mercey}\:{mr}\:{mindispower}.. \\ $$