Question Number 159534 by HongKing last updated on 18/Nov/21
$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\mathrm{arctan}^{\mathrm{2}} \:\left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$ \\ $$
Answered by chhaythean last updated on 18/Nov/21
![let { ((u=arctan^2 (x)⇒du=((2arctan(x))/(1+x^2 ))dx)),((dv=1dx⇒v=x)) :} Ω=[x∙arctan^2 (x)]_0 ^1 −2∫_0 ^1 ((x∙arctan(x))/(1+x^2 ))dx =(π^2 /(16))−2∫_0 ^1 ((x∙arctan(x))/(1+x^2 ))dx •Solve for ∫_0 ^1 ((x∙arctan(x))/(1+x^2 ))dx let x=tanθ⇒dx=sec^2 θdθ =∫_0 ^(π/4) (θ∙tanθ)dθ let { ((u=θ⇒du=dθ)),((dv=tanθdθ⇒v=−ln∣cosθ∣)) :} =[−θ∙ln∣cosθ∣]_0 ^(π/4) +∫_0 ^(π/4) ln∣cosθ∣dθ =[θ∙ln∣secθ∣]_0 ^(π/4) +∫_0 ^(π/4) ln∣cosθ∣dθ =(π/8)ln2+∫_0 ^(π/4) ln∣cosθ∣dθ •Solve for ∫_0 ^(π/4) ln∣cosθ∣dθ =∫_(π/4) ^(π/2) ln∣sinθ∣dθ=∫_(π/4) ^(π/2) (−ln2−Σ_(k=1) ^∞ ((cos(2kθ))/k))dθ =−(π/4)ln2−Σ_(k=1) ^∞ (1/k)∫_(π/4) ^(π/2) cos(2kθ)dθ =−(π/4)ln2−Σ_(k=1) ^∞ (1/k).((sin(kθ)−sin(((kπ)/2)))/(2k)) =−(π/4)ln2−(1/2)Σ_(k=1) ^∞ ((sin(((kπ)/2)))/k^2 ) =−(π/4)ln2+(1/2)Σ_(k=0) ^∞ ((sin(kπ+(π/2)))/((2k+1)^2 ))=−(π/4)ln2+(1/2)Σ_(k=0) ^∞ (((−1)^k )/((2k+1)^2 )) =−(π/4)ln2+(G/2) therefore: ∫_0 ^1 ((x∙arctan(x))/(1+x^2 ))dx=(π/8)ln2−(π/4)ln2+(G/2)=−(π/8)ln2+(G/2) We get:Ω=(π^2 /(16))−2(−(π/8)ln2+(G/2)) So determinant (((Ω=(π^2 /(16))+(π/4)ln2−G)))](https://www.tinkutara.com/question/Q159569.png)
$$\mathrm{let\begin{cases}{\mathrm{u}=\mathrm{arctan}^{\mathrm{2}} \left(\mathrm{x}\right)\Rightarrow\mathrm{du}=\frac{\mathrm{2arctan}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}}\\{\mathrm{dv}=\mathrm{1dx}\Rightarrow\mathrm{v}=\mathrm{x}}\end{cases}} \\ $$$$\Omega=\left[\mathrm{x}\centerdot\mathrm{arctan}^{\mathrm{2}} \left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}\centerdot\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:\:=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}\centerdot\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\bullet\mathrm{Solve}\:\mathrm{for}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}\centerdot\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{tan}\theta\Rightarrow\mathrm{dx}=\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\theta\centerdot\mathrm{tan}\theta\right)\mathrm{d}\theta \\ $$$$\mathrm{let\begin{cases}{\mathrm{u}=\theta\Rightarrow\mathrm{du}=\mathrm{d}\theta}\\{\mathrm{dv}=\mathrm{tan}\theta\mathrm{d}\theta\Rightarrow\mathrm{v}=−\mathrm{ln}\mid\mathrm{cos}\theta\mid}\end{cases}} \\ $$$$=\left[−\theta\centerdot\mathrm{ln}\mid\mathrm{cos}\theta\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\mid\mathrm{cos}\theta\mid\mathrm{d}\theta \\ $$$$=\left[\theta\centerdot\mathrm{ln}\mid\mathrm{sec}\theta\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\mid\mathrm{cos}\theta\mid\mathrm{d}\theta \\ $$$$=\frac{\pi}{\mathrm{8}}\mathrm{ln2}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\mid\mathrm{cos}\theta\mid\mathrm{d}\theta \\ $$$$\bullet\mathrm{Solve}\:\mathrm{for}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\mid\mathrm{cos}\theta\mid\mathrm{d}\theta \\ $$$$=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\mid\mathrm{sin}\theta\mid\mathrm{d}\theta=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \left(−\mathrm{ln2}−\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{cos}\left(\mathrm{2k}\theta\right)}{\mathrm{k}}\right)\mathrm{d}\theta \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}−\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{2k}\theta\right)\mathrm{d}\theta \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}−\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}.\frac{\mathrm{sin}\left(\mathrm{k}\theta\right)−\mathrm{sin}\left(\frac{\mathrm{k}\pi}{\mathrm{2}}\right)}{\mathrm{2k}} \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\left(\frac{\mathrm{k}\pi}{\mathrm{2}}\right)}{\mathrm{k}^{\mathrm{2}} } \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\left(\mathrm{k}\pi+\frac{\pi}{\mathrm{2}}\right)}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\pi}{\mathrm{4}}\mathrm{ln2}+\frac{\mathrm{G}}{\mathrm{2}} \\ $$$$\mathrm{therefore}:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}\centerdot\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\frac{\pi}{\mathrm{8}}\mathrm{ln2}−\frac{\pi}{\mathrm{4}}\mathrm{ln2}+\frac{\mathrm{G}}{\mathrm{2}}=−\frac{\pi}{\mathrm{8}}\mathrm{ln2}+\frac{\mathrm{G}}{\mathrm{2}} \\ $$$$\mathrm{We}\:\mathrm{get}:\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\mathrm{2}\left(−\frac{\pi}{\mathrm{8}}\mathrm{ln2}+\frac{\mathrm{G}}{\mathrm{2}}\right) \\ $$$$\mathrm{So}\:\begin{array}{|c|}{\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{4}}\mathrm{ln2}−\mathrm{G}}\\\hline\end{array} \\ $$
Commented by HongKing last updated on 18/Nov/21
$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by chhaythean last updated on 19/Nov/21
$$\mathrm{thank}\:\mathrm{you},\mathrm{sir} \\ $$