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Question Number 128245 by mnjuly1970 last updated on 05/Jan/21

                  nice  calculus...        prove  that :: ∫_0 ^( (π/4)) xcot(x)dx=(1/2)(G+(π/4)log(2))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\:::\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {xcot}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\left({G}+\frac{\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)\right) \\ $$

Answered by Dwaipayan Shikari last updated on 05/Jan/21

∫_0 ^(π/4) xcotxdx=[xlog(sinx)]_0 ^(π/4) −∫_0 ^(π/4) log(sinx)dx      = −(π/8)log(2)+(π/4)log(2)+(G/2)log(2)  (Has been proved earliar)      = (π/8)log(2)+(G/2)log(2)  Q128244

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {xcotxdx}=\left[{xlog}\left({sinx}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({sinx}\right){dx} \\ $$$$\:\:\:\:=\:−\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)+\frac{{G}}{\mathrm{2}}{log}\left(\mathrm{2}\right)\:\:\left({Has}\:{been}\:{proved}\:{earliar}\right) \\ $$$$\:\:\:\:=\:\frac{\pi}{\mathrm{8}}{log}\left(\mathrm{2}\right)+\frac{{G}}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\ $$$${Q}\mathrm{128244} \\ $$

Commented by mnjuly1970 last updated on 05/Jan/21

grateful mr payan...mercey...

$${grateful}\:{mr}\:{payan}...{mercey}... \\ $$

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