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Question Number 121869 by mnjuly1970 last updated on 12/Nov/20

$$...advancedcalculus...$$$$evaluate::$$$$\mathrm{I}\stackrel{???}{=}{\int }_0^{\frac{\pi }{2}}{\left(\frac{ln\left(tan\left(x\right)\right)}{sin\left(x\right)-cos\left(x\right)}\right)}^{2}dx$$$$.m.n.$$

Answered by mindispower last updated on 12/Nov/20

$$problemin\pi /4orPv..?$$

Commented by mnjuly1970 last updated on 13/Nov/20

$$0to\frac{\pi }{2}iscorrect...$$

Answered by mindispower last updated on 13/Nov/20

$${(sin\left(x\right)-cos\left(x\right))}^2=\frac{{(tg\left(x\right)-1)}^2}{1+{tg}^2\left(x\right)}$$$$I={\int }_0^{\frac{\pi }{2}}\frac{{ln}^2\left(tg\left(x\right)\right)}{{(tg\left(x\right)-1)}^2}dtg\left(x\right)$$$$tg\left(x\right)=t\Rightarrow I={\int }_0^{\mathrm{\infty }}\frac{{ln}^2\left(r\right)}{{(r-1)}^2}dr$$$$={\int }_0^1(\frac{{ln}^2\left(r\right)}{{(r-1)}^2}.dr+\frac{{ln}^2\left(\frac{1}{r}\right)}{{(\frac{1}{r}-1)}^2}.\frac{dr}{r^2})$$$${\int }_0^1\frac{{ln}^2\left(r\right)dr}{{(r-1)}^2}..IBP\Rightarrow$$$$\underset{t\to 0}{\lim }{\left[\frac{-{ln}^2\left(r\right)}{r-1}\right]}_t^1+2{\int }_t^1\frac{ln\left(r\right)}{r(r-1)}dr$$$$=\underset{t\to 0}{\lim }\frac{{ln}^2\left(t\right)}{t-1}+2{\int }_t^1\frac{ln\left(r\right)}{r-1}dr-2{\int }_t^1\frac{ln\left(r\right)}{r}dr$$$$=\underset{t\to 0}{\lim }\frac{{ln}^2\left(t\right)}{t-1}-2{\int }_0^{1-t}\frac{ln(1-r)}{r}dr+{ln}^2\left(t\right)$$$$=\underset{t\to 0}{\lim }[(\frac{{ln}^2\left(t\right)}{t-1}+{ln}^2\left(t\right))+2{Li}_2(1-t)]$$$$=\underset{t\to 0}{\lim }\left[\frac{{tln}^2\left(t\right)}{t-1}\right]+2{Li}_2\left(1\right)cause{Li}_{2}iscontinus$$$$=2{Li}_2\left(1\right)=2.\frac{{\pi }^2}{6}=\frac{{\pi }^2}{3}$$$$$$$$$$

Commented by mnjuly1970 last updated on 13/Nov/20

$$bravobravo$$$$sirmindspower$$$$thankyousomuch...$$

Commented by mindispower last updated on 13/Nov/20

$$thanxalwaysapleasur$$

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