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This question already has an answer here:

$$I = \int_0^1\dfrac{t \ln t}{\sqrt{1-t^2}}\mathrm{dt}$$

Attempt:

Let $t = \sin x$

$\implies dt = \cos x dx$

$\implies I = \displaystyle \int_0^{\pi/2} \dfrac{\sin x \ln (\sin x) \cos x dx}{\cos x}$

$\implies I = -\ln (\sin x) \cos x|_{0}^{\pi/2} + \displaystyle\int_0^{\pi/2}\dfrac{\cos^2 x}{\sin x} dx$

The second integral can be "evaluated" using $\cos^2 x = 1- \sin^2 x$. But unfortunately, there are finally two divergent integrals.

$\implies I = -\ln (\sin x) \cos x|_{0}^{\pi/2} + \displaystyle\int_0^{\pi/2}\csc x dx$ + $\displaystyle\int_0^{\pi/2}\sin x dx$

How do I handle this?

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marked as duplicate by Namaste integration Oct 15 '18 at 10:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Substitute $x=t^2$ in the original integral and integrate by parts. $\endgroup$ – Kemono Chen Oct 15 '18 at 5:21
  • $\begingroup$ Exactly as the previous comment says, no need for the cos change of vars. $\endgroup$ – Hashimoto Oct 15 '18 at 5:27
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    $\begingroup$ I think the integral is convergent. $\endgroup$ – kmitov Oct 15 '18 at 5:42
  • $\begingroup$ The integral is divergent. BTW, actually $I=\lim_{a\to0}-\ln\sin x\cos x|_a^{\pi/2}+\int_a^{\pi/2}\csc xdx+\int_0^{\pi/2}\sin xdx$ $\endgroup$ – Kemono Chen Oct 15 '18 at 5:45
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If you want to avoid divergent integrals, use $1-\cos x$ as the antiderivative of $\sin x$. $$I=\ln\sin x(1-\cos x)\Big|_0^{\pi/2}+\int_0^{\pi/2}(\cos x-1)\cot xdx\\ =0+\ln2-1=\ln2-1.$$

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  • $\begingroup$ How can we use 1- cos x as the antiderivative $\endgroup$ – Abcd Oct 15 '18 at 6:25
  • $\begingroup$ Note that you used $\int_0^{\pi/2} \sin x\ln\sin xdx=\int\sin xdx \ln\sin x|_0^{\pi/2}-\int_0^{\pi/2}(\int \sin xdx)(\ln \sin x)'dx$. $\int \sin xdx=C-\cos x$. You should choose a suitable $C$ to avoid divergency. $\endgroup$ – Kemono Chen Oct 15 '18 at 6:29
  • $\begingroup$ I have never seen integration being done that way. $\endgroup$ – Abcd Oct 15 '18 at 6:30
  • $\begingroup$ It is just a fancy notation of integrating by parts. $\endgroup$ – Kemono Chen Oct 15 '18 at 6:32
  • $\begingroup$ What method did you use to evaluate the second integral? $\endgroup$ – Abcd Oct 15 '18 at 6:48
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I would just focus on the antiderivative.

Using you substitution and continuing with the tangent half-angle substitution, we should get $$I=\int \dfrac{\sin (x) \ln (\sin (x)) \cos(x) }{\cos x}\,dx=\int\sin (x) \ln (\sin (x))\,dx$$ and integration by parts would lead to $$I=\cos (x)+\log \left(\sin \left(\frac{x}{2}\right)\right)-\log \left(\cos \left(\frac{x}{2}\right)\right)-\cos (x) \log (\sin (x))$$ Now, looking at the Taylor expansion around $x=0$, $$I=(1-\log (2))+x^2 \left(\frac{\log (x)}{2}-\frac{1}{4}\right)+O\left(x^4\right)$$ and around $x=\frac \pi 2$ $$I=-\frac{1}{6} \left(x-\frac{\pi }{2}\right)^3+O\left(\left(x-\frac{\pi }{2}\right)^4\right)$$ and then the result already given by @Kemono Chen.

Edit

Back to $t$, $$\int\dfrac{t \ln t}{\sqrt{1-t^2}}\,{dt}=\sqrt{1-t^2} (1-\log (t))+\log \left(\frac{t}{1+\sqrt{1-t^2}}\right)$$

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