Problem 33

Determine \displaystyle \int_b^a \frac 1 {\sqrt {(a - x)(x - b)}} \mathrm dx.

[originally posted by Benjy100 on TSR]

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For brevity write:

\displaystyle I = \int_b^a \frac 1 {\sqrt{(a-x)(x-b)}} \mathrm dx

Use the substitution x = a \sin^2 \theta + b \cos^2 \theta, then \mathrm dx = (2 a \sin \theta \cos \theta - 2 b \sin \theta \cos \theta) \mathrm d\theta = 2(a - b) \sin \theta \cos \theta. Then:

\begin{align*}I & = 2 (a - b)\int_0^{\pi/2} \frac {\sin \theta \cos \theta} {\sqrt{(a - a \sin^2 \theta - b \cos^2 \theta) (b \cos^2 \theta + a \sin^2 \theta - b)}} \mathrm d\theta \\ & = 2(a - b) \int_0^{\pi/2} \frac {\sin \theta \cos \theta} {\sqrt {(a \cos^2 \theta - b \cos^2 \theta)(a \sin^2 \theta -b \sin^2 \theta)}} \mathrm d\theta\end{align*}

using \sin^2 \theta + \cos^2 \theta \equiv 1 to get a(1 - \sin^2 \theta) = a \cos^2 \theta and b(\cos^2 \theta - 1) = -b(1 - \cos^2 \theta) = -b \sin^2 \theta.

Continuing:

\begin{align*}I & = 2(a - b) \int_0^{\pi/2} \frac {\sin \theta \cos \theta} {\sqrt {(a - b)} \sqrt {\cos^2 \theta} \sqrt {(a - b)} \sqrt {\sin^2 \theta}} \mathrm d\theta \\ & = 2 \int_0^{\pi/2} \frac {\sin \theta \cos \theta} {\sin \theta \cos \theta} \mathrm d\theta \\ & = 2 \int_0^{\pi/2} \mathrm d\theta \\ & = \pi\end{align*}

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Fantastic content!!! :clap: Amazing that it generalises like that

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