<s>[INST] How to integrate $\int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \,dt$? How to integrate $\displaystyle \int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \,dt$?
I literally have no idea how to integrate this integral. 
I've tried all basic methods, but it seems like a quite hard problem for a beginner.
 [/INST] Do the following substitution 
$$t=a\cos^2 \theta+b \sin^2 \theta \implies \mathrm dt=(b-a) \sin 2 \theta \, \mathrm d \theta$$
Your function $\displaystyle \sqrt{\frac{t - a}{b - t}}$ will be tremendously simplified to $\tan \theta$. Hence, your integral will be simplified :
\begin{align}
\int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \, \mathrm dt&=\int_0^{{\pi}/{2}} (a\sin^2 \theta+b \cos^2 \theta) \cdot  \tan \theta \,((b-a) \sin 2 \theta \,  \mathrm d \theta)\\
&= ~a\int_0^{{\pi}/{2}} \frac{\sin^3 \theta}{\cos \theta}\, \cdot (b-a) \sin 2 \theta \, \mathrm d \theta+ b\int_0^{{\pi}/{2}}\frac{ \sin 2 \theta}{2} \,  (b-a) \sin 2 \theta \, \mathrm d \theta\\
&=~(b-a) \left(2a \int_0^{{\pi}/{2}} {\sin^4 \theta}\, \mathrm d \theta + \frac b2 \int_0^{{\pi}/{2}}{ \sin^2 2 \theta} \, \mathrm d \theta \right)\\
&=~(b-a) \left(2a \cdot\frac{3\pi}{16} + \frac b2 \cdot\frac{\pi}{4}\right) \\
&=\frac\pi8 (b-a) (3a+b)
\end{align}
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