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Background(optional info): This comes from an example (3.3) of Feynman's trick or differentiating under the integral sign by parameter substitution. The integral $-\int_0^{\pi/2}x\cot(x)$dx is evaluated ($=-\frac{\pi}{2}\log(2)$) by adding a parameter b, modifying the integrand, and switching the integral and differential $$\frac{\text{d}}{\text{db}}I(b)=\int_0^{\pi/2}\frac{\partial}{\partial{b}}\frac{\tan^{-1}(b\tan(x))}{\tan(x)}dx$$ then taking the $\lim_{b\to 1}$ after integrating w.r.t b.


Anyway, my question is at the end it is stated that this is another way to integrate $\int_0^{\pi/2}\log(\sin(x))\text{dx}$ because $$\int_0^{\pi/2}\log(\sin(x))\text{dx}=-\int_0^{\pi/2}x\cot(x)dx \tag{1}$$

How do you convert the first into the second integral in (1)?

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    $\begingroup$ Integration by parts. $\endgroup$ – Lord Shark the Unknown Jun 4 '17 at 6:59
  • $\begingroup$ @LordSharktheUnknown I mean, how do you convert the integrand of the first into the second, not how to solve it. $\endgroup$ – user5389726598465 Jun 4 '17 at 7:00
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    $\begingroup$ Integration by parts. $\endgroup$ – Lord Shark the Unknown Jun 4 '17 at 7:00
  • $\begingroup$ @LordSharktheUnknown Aha! I see it now. $\endgroup$ – user5389726598465 Jun 4 '17 at 7:03

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