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Compute $\int_{0}^{\pi/2}\cos(x)\ln(\tan(x))dx$

It is easy to check this improper integral converges. One also notes that $\int_{0}^{\pi/2}\cos(x)\ln(\tan(x))dx=-\int_{0}^{\pi/2}\sin(x)\ln(\tan(x))dx$.

It is possible to find an antiderivative for $\cos(x)\ln(\tan(x))$ using integration by parts, but I'm looking for a nicer way, that would involve only substitution or other tricks.

Gabriel Romon
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    Downvoter ? ${}{}$ – Gabriel Romon May 15 '15 at 04:42
  • I'm not the downvoter, but my guess would be that your question was downvoted on grounds of being a duplicate (though if that's the case, link to duplicate question really should have been provided). – David H May 15 '15 at 04:48

1 Answers1

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For $x \in [0,\frac{\pi}{2}]$, we have that $\cos x = \sqrt{1 - \sin^2 x}$, so $$ \log (\tan x) = \log (\sin x) - \log (\cos x) = \log (\sin x) - \frac{1}{2} \log ( 1 - \sin^2 x)$$ Then make the substitution $u = \sin x$. Since $du = \cos x \;dx$, we get that $$\int_{0}^{\pi/2}\cos(x)\log(\tan(x))dx$$ $$\int_{0}^{\pi/2}\cos(x)\log(\sin(x))dx-\frac{1}{2} \int_{0}^{\pi/2}\cos(x)\log(1-\sin^2 x)dx$$ $$\int_0^1 \log u\; du -\frac{1}{2} \int_0^1 \log(1-u^2)\; du $$$$= \int_0^1 \log u \;du -\frac{1}{2} \int_0^1 \log(1-u)+\log(1+u)\; du$$ and this can be easily integrated, since the antiderivative of $\log y$ is equal to $y \log y -y +C$.

shalop
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