Question Number 30184 by abdo imad last updated on 17/Feb/18
$${find}\:\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{2}} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctanx}\:{dx}\:.\:\left({arctan}={tan}^{−\mathrm{1}} \right). \\ $$
Commented by abdo imad last updated on 20/Feb/18
![the ch.x=(1/t) give I = ∫_(1/2) ^2 (1+t^2 )arctan((1/t))(dt/t^2 ) =∫_(1/2) ^2 (1+(1/t^2 ))((π/2) −arctant)dt =(π/2) ∫_(1/2) ^2 (1+(1/t^2 ))dt −∫_(1/2) ^2 (1+(1/t^2 ))arctantdt⇒2I=(π/2)∫_(1/2) ^2 (1+(1/t^2 ))dt 2I= (π/2)(3/2) +(π/2)[−(1/t)]_(1/2) ^2 = ((3π)/4) +(π/2)(2 −(1/2))=((3π)/4) +((3π)/4) =((3π)/2) ⇒ I= ((3π)/4) .](https://www.tinkutara.com/question/Q30344.png)
$${the}\:{ch}.{x}=\frac{\mathrm{1}}{{t}}\:{give}\:{I}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right){arctan}\left(\frac{\mathrm{1}}{{t}}\right)\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)\left(\frac{\pi}{\mathrm{2}}\:−{arctant}\right){dt} \\ $$$$=\frac{\pi}{\mathrm{2}}\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}\:−\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){arctantdt}\Rightarrow\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt} \\ $$$$\mathrm{2}{I}=\:\frac{\pi}{\mathrm{2}}\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}\left[−\frac{\mathrm{1}}{{t}}\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{2}} =\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:+\frac{\pi}{\mathrm{2}}\left(\mathrm{2}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{3}\pi}{\mathrm{4}}\:+\frac{\mathrm{3}\pi}{\mathrm{4}}\:=\frac{\mathrm{3}\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${I}=\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:. \\ $$