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.
 A: $ \frac{t-a}{b-t} $ increases from $0$ to $\infty$ on the interval of integration, so let's try $u^2=\frac{t-a}{b-t}$. Then $t = \frac{a+bu^2}{1+u^2} $, so $dt = \frac{2(b-a)u}{(1+u^2)^2}$. Hence the integral becomes
$$ 2(b-a)\int_0^{\infty} \frac{u^2(a+bu^2)}{(1+u^2)^3} \, du, $$
which can be done using, for example, integration by parts.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\left.\int_{a}^{b}t\root{t - a \over b - t}\,\dd t\,\right\vert_{\ a\ <\ b} &
\stackrel{t\ \mapsto\ t - \pars{a + b}/2}{=}\,\,\,
\int_{-\pars{b - a}/2}^{\pars{b - a}/2}\pars{t + {a + b \over 2}}\root{t + \pars{b - a}/2 \over \pars{b - a}/2 - t}\,\dd t
\\[5mm] &
\stackrel{2t/\pars{b - a}\ \mapsto\ t}{=}\,\,\,
{b - a \over 2}\bracks{{b - a \over 2}\!\int_{-1}^{1}\!t\root{1 + t \over 1 - t}\,\dd t +
{b + a \over 2}\!\int_{-1}^{1}\!\root{1 + t \over 1 - t}\,\dd t}
\\[5mm] & =
{\pars{b - a}^{2} \over 4}\int_{-1}^{1}{t\root{1 - t^{2}} \over 1 - t}\,\dd t +
{b^{2} - a^{2} \over 4}\int_{-1}^{1}{\root{1 - t^{2}} \over 1 - t}\,\dd t
\\[5mm] & =
{\pars{b - a}^{2} \over 2}\
\underbrace{\int_{0}^{1}{t^{2} \over \root{1 - t^{2}}}\,\dd t}
_{\ds{\pi \over 4}}\ +\
{b^{2} - a^{2} \over 2}\
\underbrace{\int_{0}^{1}{\dd t \over \root{1 - t^{2}}}}_{\ds{\pi \over 2}}
\end{align}

The last integrals were evaluated with the substitution $\ds{t \equiv \sin\pars{\theta}}$.


$$
\left.\int_{a}^{b}t\root{t - a \over b - t}\,\dd t\,\right\vert_{\ a\ <\ b} =
\bbx{{\pars{b - a}\pars{3b + a} \over 8}\,\pi}
$$
A: 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}
A: HINT:
WLOG $b\ge a$
For real calculus, we need $b>t\ge a$
$$\iff b-\dfrac{a+b}2>t-\dfrac{a+b}2\ge a-\dfrac{a+b}2\iff\dfrac{b-a}2>t-\dfrac{a+b}2\ge-\dfrac{b-a}2$$
WLOG we can choose $t-\dfrac{a+b}2=\dfrac{b-a}2\cos2y$ where $0\le2y\le\pi$
$$\implies dt=(b-a)\sin2y\ dy$$
and $2t=a+b+(b-a)\cos2y=2(b\cos^2y+a\sin^2y)$
$$\dfrac{t-a}{b-t}=\dfrac{(b-a)\cos^2y}{(b-a)\sin^2y}$$
As  $0\le2y\le\pi,$ $$\sqrt{\dfrac{t-a}{b-t}}=+\cot y$$
