Question Number 33328 by prof Abdo imad last updated on 14/Apr/18
$${find}\:\:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctanx}\:{dx} \\ $$
Commented by math khazana by abdo last updated on 19/Apr/18
![let put I = ∫_(π/4) ^(4/π) (1+(1/x^2 ))arctanx dx .let integrate by parts u^′ =1+(1/x^2 ) and v =arctanx I = [(1−(1/x))arctanx]_(π/4) ^(4/π) −∫_(π/4) ^(4/π) (1−(1/x)) (dx/(1+x^2 )) = (1−(π/4))arctan((4/π)) −(1−(4/π)) −∫_(π/4) ^(4/π) (dx/(1+x^2 )) + ∫_(π/4) ^(4/π) (dx/(x( 1+x^2 ))) but ∫_(π/4) ^(4/π) (dx/(1+x^2 )) = arctan( (4/π)) −arctan((π/4)) =(π/2) −1−1=(π/2) −2 let?decompose F(x) = (1/(x(1+x^2 ))) = (a/x) +((bx +c)/(1+x^2 )) a =lim_(x→0) x F(x) = 1 lim_(x→+∞) x F(x) =0 = a +b ⇒b=−a =−1 F(x) = (1/x) +((−x +c)/(1+x^2 )) we look tbat c=0 ⇒ F(x) = (1/x) −(x/(1+x^2 )) ⇒ ∫_(π/4) ^(4/π) (dx/(x(1+x^2 ))) = ∫_(π/4) ^(4/π) ((1/x) −(x/(1+x^2 )))dx =[ ln(x)−(1/2)ln(1+x^2 )]_(π/4) ^(4/π) =[ln((x/( (√(1+x^2 )))))]_(π/4) ^(4/π) = ln( (4/(π(√(1+((16)/π^2 )))))) −ln( (π/(4(√(1+(π^2 /(16))))))) I =(1−(π/4))((π/2) −1) +1 +(4/π) −(π/2) +ln( (4/(π(√(1+((16)/π^2 )))))) −ln( (π/(4(√(1+(π^2 /(16))))))) .](https://www.tinkutara.com/question/Q33566.png)
$${let}\:{put}\:{I}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctanx}\:{dx}\:.{let}\:{integrate}\:{by} \\ $$$${parts}\:{u}^{'} \:=\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\:{and}\:{v}\:={arctanx} \\ $$$${I}\:=\:\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){arctanx}\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:−\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\:\left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right){arctan}\left(\frac{\mathrm{4}}{\pi}\right)\:−\left(\mathrm{1}−\frac{\mathrm{4}}{\pi}\right)\:−\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$+\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\:\:\:\:\frac{{dx}}{{x}\left(\:\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\:{but} \\ $$$$\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:{arctan}\left(\:\frac{\mathrm{4}}{\pi}\right)\:−{arctan}\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\:−\mathrm{1}−\mathrm{1}=\frac{\pi}{\mathrm{2}}\:−\mathrm{2}\:\:{let}?{decompose} \\ $$$${F}\left({x}\right)\:=\:\:\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\:\frac{{a}}{{x}}\:\:+\frac{{bx}\:+{c}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{0}} {x}\:{F}\left({x}\right)\:=\:\mathrm{1} \\ $$$${lim}_{{x}\rightarrow+\infty} {x}\:{F}\left({x}\right)\:=\mathrm{0}\:=\:{a}\:+{b}\:\Rightarrow{b}=−{a}\:=−\mathrm{1} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}\:\:+\frac{−{x}\:+{c}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\:\:{we}\:{look}\:{tbat}\:{c}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}\:\:−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:\left(\frac{\mathrm{1}}{{x}}\:−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$$$=\left[\:{ln}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \:=\left[{ln}\left(\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{4}}{\pi}} \\ $$$$=\:{ln}\left(\:\:\frac{\mathrm{4}}{\pi\sqrt{\mathrm{1}+\frac{\mathrm{16}}{\pi^{\mathrm{2}} }}}\right)\:−{ln}\left(\:\:\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{16}}}}\right) \\ $$$${I}\:=\left(\mathrm{1}−\frac{\pi}{\mathrm{4}}\right)\left(\frac{\pi}{\mathrm{2}}\:−\mathrm{1}\right)\:+\mathrm{1}\:+\frac{\mathrm{4}}{\pi}\:−\frac{\pi}{\mathrm{2}}\: \\ $$$$+{ln}\left(\:\frac{\mathrm{4}}{\pi\sqrt{\mathrm{1}+\frac{\mathrm{16}}{\pi^{\mathrm{2}} }}}\right)\:−{ln}\left(\:\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{16}}}}\right)\:. \\ $$