How to integrate $\int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \,dt$?

Question

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.

Answer 1

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}

Answer 2

$ \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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{t \equiv \sin\pars{\theta}}$.

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$$ \left.\int_{a}^{b}t\root{t - a \over b - t}\,\dd t\,\right\vert_{\ a\ <\ b} = \bbx{{\pars{b - a}\pars{3b + a} \over 8}\,\pi} $$

Answer 4

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 $0\le2y\le\pi,$ $$\sqrt{\dfrac{t-a}{b-t}}=+\cot y$$