Find-0-1-arctan-2-x-dx-

Question Number 159534 by HongKing last updated on 18/Nov/21

Find :

Ω = \underset{0}{\int^{1}} \arctan^{2} (x) dx

Answered by chhaythean last updated on 18/Nov/21

let {\begin{cases} u = \arctan^{2} (x) \Rightarrow du = \frac{2 \arctan (x)}{1 + x^{2}} dx \\ dv = 1 dx \Rightarrow v = x \end{cases}

Ω = {[x \cdot \arctan^{2} (x)]}_{0}^{1} - 2 \int_{0}^{1} \frac{x \cdot \arctan (x)}{1 + x^{2}} dx

= \frac{π^{2}}{16} - 2 \int_{0}^{1} \frac{x \cdot \arctan (x)}{1 + x^{2}} dx

∙ Solve for \int_{0}^{1} \frac{x \cdot \arctan (x)}{1 + x^{2}} dx

let x = \tan θ \Rightarrow dx = \sec^{2} θ d θ

= \int_{0}^{\frac{π}{4}} (θ \cdot \tan θ) d θ

let {\begin{cases} u = θ \Rightarrow du = d θ \\ dv = \tan θ d θ \Rightarrow v = - \ln ∣ \cos θ ∣ \end{cases}

= {[- θ \cdot \ln ∣ \cos θ ∣]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} \ln ∣ \cos θ ∣ d θ

= {[θ \cdot \ln ∣ \sec θ ∣]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} \ln ∣ \cos θ ∣ d θ

= \frac{π}{8} \ln 2 + \int_{0}^{\frac{π}{4}} \ln ∣ \cos θ ∣ d θ

∙ Solve for \int_{0}^{\frac{π}{4}} \ln ∣ \cos θ ∣ d θ

= \int_{\frac{π}{4}}^{\frac{π}{2}} \ln ∣ \sin θ ∣ d θ = \int_{\frac{π}{4}}^{\frac{π}{2}} (- \ln 2 - \underset{k = 1}{\sum^{\infty}} \frac{\cos (2 k θ)}{k}) d θ

= - \frac{π}{4} \ln 2 - \underset{k = 1}{\sum^{\infty}} \frac{1}{k} \int_{\frac{π}{4}}^{\frac{π}{2}} \cos (2 k θ) d θ

= - \frac{π}{4} \ln 2 - \underset{k = 1}{\sum^{\infty}} \frac{1}{k} . \frac{\sin (k θ) - \sin (\frac{k π}{2})}{2 k}

= - \frac{π}{4} \ln 2 - \frac{1}{2} \underset{k = 1}{\sum^{\infty}} \frac{\sin (\frac{k π}{2})}{k^{2}}

= - \frac{π}{4} \ln 2 + \frac{1}{2} \underset{k = 0}{\sum^{\infty}} \frac{\sin (k π + \frac{π}{2})}{{(2 k + 1)}^{2}} = - \frac{π}{4} \ln 2 + \frac{1}{2} \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}}

= - \frac{π}{4} \ln 2 + \frac{G}{2}

therefore : \int_{0}^{1} \frac{x \cdot \arctan (x)}{1 + x^{2}} dx = \frac{π}{8} \ln 2 - \frac{π}{4} \ln 2 + \frac{G}{2} = - \frac{π}{8} \ln 2 + \frac{G}{2}

We get : Ω = \frac{π^{2}}{16} - 2 (- \frac{π}{8} \ln 2 + \frac{G}{2})

So \begin{matrix} Ω = \frac{π^{2}}{16} + \frac{π}{4} \ln 2 - G \end{matrix}

Commented by HongKing last updated on 18/Nov/21

perfect my dear Ser thank you so much

Commented by chhaythean last updated on 19/Nov/21

thank you, sir