Integrate $\int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{\tan x} \,dx}$ 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?
 A: 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 $(1)$.
A: 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 $u=\cos x$. 
A: Using $$\tan(x)\cos(x)\sin(x)=\sin^2(x)=\frac{\tan^2(x)}{1+\tan^2(x)}$$ and putting $t=\tan(x)$, we get
$$I=\int_0^{+\infty}t^{\frac{3}{2}}dt=+\infty.$$
