Question Number 96280 by bobhans last updated on 31/May/20
Commented by bemath last updated on 31/May/20
![let w = arctan x dw = (dx/(1+x^2 )) ⇒ ∫ w dw = [(1/2)w^2 ]_0 ^(π/2) = (1/2)((π/2))^2 = (π^2 /8) .](https://www.tinkutara.com/question/Q96304.png)
$$\mathrm{let}\:\mathrm{w}\:=\:\mathrm{arctan}\:\mathrm{x} \\ $$$$\mathrm{dw}\:=\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\:\int\:\mathrm{w}\:\mathrm{dw}\:=\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}^{\mathrm{2}} \:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} =\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:. \\ $$
Answered by 675480065 last updated on 31/May/20
![let u=arctanx ⇒tanu=x ⇒sec^2 udu=dx ⇒(1+tan^2 u)du=dx {but tanu=x} ⇒(1+x^2 )du=dx ⇒∫_0 ^∞ (((arctanx)/(x^2 +1)))dx=∫_(0 ) ^∞ ((((x^2 +1)udu)/(x^2 +1))) =∫_0 ^∞ udu [((tan^(−2) x)/2)]_0 ^∞ =(π^2 /8)](https://www.tinkutara.com/question/Q96283.png)
$$\mathrm{let}\:\mathrm{u}=\mathrm{arctanx} \\ $$$$\Rightarrow\mathrm{tanu}=\mathrm{x} \\ $$$$\Rightarrow\mathrm{sec}^{\mathrm{2}} \mathrm{udu}=\mathrm{dx} \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{u}\right)\mathrm{du}=\mathrm{dx}\:\:\left\{\mathrm{but}\:\mathrm{tanu}=\mathrm{x}\right\} \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{du}=\mathrm{dx} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{arctanx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{dx}=\int_{\mathrm{0}\:} ^{\infty} \left(\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{udu}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \mathrm{udu} \\ $$$$\left[\frac{\mathrm{tan}^{−\mathrm{2}} \mathrm{x}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\infty} =\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Commented by bobhans last updated on 31/May/20
$$\mathrm{good}.\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mathmax by abdo last updated on 31/May/20
![A =∫_0 ^∞ ((arctan(x))/(x^2 +1))dx by parts u^′ =(1/(x^2 +1)) and v=arctanx A =[arctan^2 x]_0 ^∞ −∫_0 ^∞ ((arctanx)/(1+x^2 ))dx =(π^2 /4) −A ⇒2A =(π^2 /4) ⇒A =(π^2 /8)](https://www.tinkutara.com/question/Q96301.png)
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{u}^{'} \:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\:\mathrm{and}\:\mathrm{v}=\mathrm{arctanx} \\ $$$$\mathrm{A}\:=\left[\mathrm{arctan}^{\mathrm{2}} \mathrm{x}\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:−\mathrm{A}\:\Rightarrow\mathrm{2A}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\mathrm{A}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$