Challenge question: solve $$\int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{\tan x} \,dx}$$
It's a generalization of a recent Math.SE question, but how would one normally approach it?
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2To which math.se question are you referencing? – user190080 Dec 30 '16 at 17:27
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2i have only found a numerical value of your integral $$0.2383627095847316199842751304939977634136721384754172600021618182991202279207338258104065665273398798$$ – Dr. Sonnhard Graubner Dec 30 '16 at 17:32
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3@user190080 : This one by JohnChessant. He has a function $I(a)$, and he is asking for $I(2)$. I simply posed $I(4)$ (this question). – THA Dec 31 '16 at 00:34
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2@THA I have been summoned. :D Yes, $I(4)$ is equivalent to the integral in this question, and by my formula the answer is $\frac{1}{16}\pi\left(4−\pi\right)\sqrt{2}$ (talk of funny answers ;) ). – Ant Dec 31 '16 at 01:08
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Hint. By the change of variable $$ \sqrt{\tan x}=t,\quad x=\arctan (t^2),\quad dx=\frac{2t\:dt}{1+t^4}, $$ one gets
$$ \int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{\tan x} \,dx}=\int_0^\infty \frac{4t^4\ln t}{\left(1+t^4\right)^2}\:dt=\frac{\pi \sqrt{2}}{4}-\frac{\pi ^2\sqrt{2}}{16}. \tag1 $$
Addendum. One may recall the identity $$ \int_0^\infty\frac{x^\alpha}{1+x}\:dx=-\frac{\pi}{\sin \alpha \pi}, \quad -1<\alpha<0, \tag2 $$ giving, with an integration by parts, $$ \int_0^\infty\frac{x^\alpha}{(1+x)^2}\:dx=\frac{\alpha \pi}{\sin \alpha \pi}, \quad -1<\alpha<1 \tag3 $$ differentiating $(3)$ yields $$ \int_0^\infty\frac{x^\alpha\ln x}{(1+x)^2}\:dx=\frac{\pi}{\sin \alpha \pi}-\frac{\pi^2\alpha \cos \alpha \pi}{\sin^2 \alpha \pi}, \quad -1<\alpha<1, \tag4 $$ then putting $x=t^4$, $\alpha=\frac14$ gives $(1)$.
Olivier Oloa
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1Did you use integration by parts in the end? If yes, Could you please explain how did u evaluate $\int{\frac{4t^4}{(1+t^4)^2}dt}$ – Serenity Dec 30 '16 at 18:06
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@ReneSchipperus I assumed he has integrated by parts, taking $lnt$ as the first function – Serenity Dec 30 '16 at 18:12
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+1 Happy new year all! And it appears there is no other way to do it; I asked the question that OP referenced; you can check that my formula was essentially equivalent to that of @OlivierOloa. – Ant Dec 31 '16 at 03:50
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\begin{align} &\int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{(\tan x)} \,dx}\\ =& - \frac12 \int_0^{\pi/2} \sqrt{\tan x}\ln (\tan x) \ d( \cos^2 x) \overset{ibp}= \frac12\int_0^{\pi/2}\frac{1+\ln \sqrt{\tan x}}{\sqrt{\tan x}}\ dx\\ =& \int_0^\infty \frac{1+\ln t}{1+t^4}dt =\frac{\pi }{2 \sqrt{2}}-\frac{\pi ^2}{8\sqrt{2}} \end{align}
Quanto
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Write $\ln\tan x=\ln\sin x-\ln\cos x$, replace $\sqrt{\tan x}$ also, and do the integral of the two summands separately. For the first summand put $u=\sin x$ and integrate by parts. For the second summand put $u=\cos x$.
user2747939
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hint
Use $$\cos(x)\sin(x)=\frac{\tan(x)}{1+\tan^2(x)}$$ and put $$t=\tan(x)$$.
hamam_Abdallah
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There was nothing modified that could have modified the question's meaning. In fact, the only edit was to include \int in the dollar signs. – amWhy Dec 30 '16 at 18:21
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If you want to delete your answer, you are free to do so. The question now is, as all long, not changed. – amWhy Dec 30 '16 at 18:22