Question Number 126971 by mnjuly1970 last updated on 25/Dec/20
$$\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:\:{evaluate}\:': \\ $$$$\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({arctan}\left({x}\right)\right)^{\mathrm{2}} {dx}=? \\ $$$$\:\:\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 25/Dec/20
![∫_0 ^(π/4) t^2 (1+x^2 )dt tan^(−1) x=t⇒(1/(1+x^2 ))=(dt/dx) =∫_0 ^(π/4) t^2 sec^2 t dt =[t^2 tant]_0 ^(π/4) −2∫_0 ^(π/4) t.tant dt =(π^2 /(16))+[2tlog(cost)]_0 ^(π/4) −2∫_0 ^(π/4) log(cost)dt =(π^2 /(16))−(π/4)log(2)−2((G/2) −(π/4)log(2)) =(π^2 /(16))+(π/4)log(2)−G G=Catalan Constant Merry Christmas 🔔🤶](Q126976.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {t}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right){dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}^{−\mathrm{1}} {x}={t}\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{{dt}}{{dx}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {t}^{\mathrm{2}} {sec}^{\mathrm{2}} {t}\:{dt}\:=\left[{t}^{\mathrm{2}} {tant}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {t}.{tant}\:{dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\left[\mathrm{2}{tlog}\left({cost}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({cost}\right){dt} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−\frac{\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\mathrm{2}\left(\frac{{G}}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−{G} \\ $$$${G}={Catalan}\:{Constant} \\ $$$${Merry}\:{Christmas} \\ $$$$ \\ $$🔔🤶
Commented by mnjuly1970 last updated on 25/Dec/20
$${grateful}\:{mr} \\ $$$${payan}\:{and} \\ $$$${merry}\:{christmas} \\ $$