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Question Number 126971 by mnjuly1970 last updated on 25/Dec/20

$$...nicecalculus...$$$$evaluate{}^{\prime }:$$$$\mathrm{\Omega }={\int }_0^1{\left(arctan\left(x\right)\right)}^{2}dx=?$$$$$$

Answered by Dwaipayan Shikari last updated on 25/Dec/20

$${\int }_0^{\frac{\pi }{4}}{t}^2(1+x^2)dt{tan}^{-1}x=t\Rightarrow \frac{1}{1+x^2}=\frac{dt}{dx}$$$$={\int }_0^{\frac{\pi }{4}}{t}^2{sec}^{2}tdt={\left[t^{2}tant\right]}_0^{\frac{\pi }{4}}-2{\int }_0^{\frac{\pi }{4}}t.tantdt$$$$=\frac{{\pi }^2}{16}+{\left[2tlog\left(cost\right)\right]}_0^{\frac{\pi }{4}}-2{\int }_0^{\frac{\pi }{4}}log\left(cost\right)dt$$$$=\frac{{\pi }^2}{16}-\frac{\pi }{4}log\left(2\right)-2(\frac{G}{2}-\frac{\pi }{4}log\left(2\right))$$$$=\frac{{\pi }^2}{16}+\frac{\pi }{4}log\left(2\right)-G$$$$G=CatalanConstant$$$$MerryChristmas$$$$$$🔔🤶

Commented by mnjuly1970 last updated on 25/Dec/20

$$gratefulmr$$$$payanand$$$$merrychristmas$$

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