Question Number 96097 by mhmd last updated on 29/May/20
Answered by mathmax by abdo last updated on 29/May/20
![A =∫_(−1) ^1 (dx/((e^x +1)(x^2 +1))) changement x=−t give A =−∫_(−1) ^1 ((−dt)/((e^(−t) +1)(t^2 +1))) =∫_(−1) ^1 (e^t /((e^t +1)(t^2 +1)))dt =∫_(−1) ^1 ((e^t +1−1)/((e^t +1)(t^2 +1)))dt =∫_(−1) ^1 (dt/(t^(2 ) +1)) −A ⇒2A =2∫_0 ^1 (dt/(1+t^2 )) ⇒ A =[arctant]_0 ^1 =(π/4)](https://www.tinkutara.com/question/Q96100.png)
$$\mathrm{A}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\left(\mathrm{e}^{\mathrm{x}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\mathrm{changement}\:\mathrm{x}=−\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{A}\:=−\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{−\mathrm{dt}}{\left(\mathrm{e}^{−\mathrm{t}} +\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{e}^{\mathrm{t}} }{\left(\mathrm{e}^{\mathrm{t}} +\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)}\mathrm{dt} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{e}^{\mathrm{t}} +\mathrm{1}−\mathrm{1}}{\left(\mathrm{e}^{\mathrm{t}} \:+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)}\mathrm{dt}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}\:} +\mathrm{1}}\:−\mathrm{A}\:\Rightarrow\mathrm{2A}\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{A}\:=\left[\mathrm{arctant}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\pi}{\mathrm{4}} \\ $$
Commented by mhmd last updated on 29/May/20
$${sir}\:{are}\:{you}\:{can}\:{exactly}\:{steb}\:{by}\:{steb}\:{pleas}\:{sir}\:? \\ $$
Commented by mhmd last updated on 29/May/20
![how change the interval intigral from[−1,1] to[0,1] can you exactly sir ? in befor last steb](https://www.tinkutara.com/question/Q96107.png)
$${how}\:{change}\:{the}\:{interval}\:{intigral}\:{from}\left[−\mathrm{1},\mathrm{1}\right]\:{to}\left[\mathrm{0},\mathrm{1}\right]\:{can}\:{you}\:{exactly}\:{sir}\:?\:{in}\:{befor}\:{last}\:{steb} \\ $$
Commented by Sourav mridha last updated on 29/May/20
![so simple if f(x)=f(−x) then f(x) is called even f^n then:∫_(−a) ^(+a) f(x)=∫_0 ^a [f(x)+f(−x)] dx =2∫_0 ^a f(x)dx for even f(x) =0for odd f(x),f(x)=−f(−x). it just property of definite integral.](https://www.tinkutara.com/question/Q96110.png)
$$\mathrm{so}\:\mathrm{simple}\:\mathrm{if}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(−\mathrm{x}\right)\:\mathrm{then}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{called}\:\mathrm{even}\:\mathrm{f}^{\mathrm{n}} \\ $$$$\mathrm{then}:\int_{−\mathrm{a}} ^{+\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\mathrm{a}} \left[\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(−\mathrm{x}\right)\right]\:\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{a}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:\mathrm{for}\:\mathrm{even}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0for}\:\mathrm{odd}\:\mathrm{f}\left(\mathrm{x}\right),\mathrm{f}\left(\mathrm{x}\right)=−\mathrm{f}\left(−\mathrm{x}\right). \\ $$$$\mathrm{it}\:\mathrm{just}\:\mathrm{property}\:\mathrm{of}\:\mathrm{definite}\:\mathrm{integral}. \\ $$