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

Answer 5

A method which is relatively general (worth trying whenever you have the square root of the quotient of two linear expressions) is to multiply top and bottom by the square root of the linear expression in the numerator. Here $$\int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \,dt = \int_a^b \frac{t(t - a)}{\sqrt{(t-a)(b - t)}} \,dt.$$ For this, the standard method is to use a trigonometric or hyperbolic substitution as appropriate for the quadratic in the square root in the denominator. Here $$\int_a^b \frac{t(t - a)}{\sqrt{(t-a)(b - t)}} \,dt = \int_a^b \frac{t(t - a)}{\sqrt{\tfrac{1}{4}(b-a)^2-(t-\tfrac{1}{2}(b+a))^2}} \,dt $$ so use $t-\tfrac{1}{2}(b+a)= \tfrac{1}{2}(b-a)\sin \theta$ and the integral becomes $$\tfrac{1}{4}(b-a)\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} (b+a)+2b\sin \theta +(b-a)\sin^2 \theta \,dt = \tfrac{1}{8}\pi(b-a)(3b+a).$$