Question

Use the substitution x = 2 − cos θ to evaluate the integral ∫ 2 3/2 ( x − 1 3 − x )1 2 dx. Show that, for a < b, ∫ q p ( x − a b − x )1 2 dx = (b − a)(π + 3√ 3 − 6) 12 , where p = ???????????????????????????

Answer 1

If the integral as written in my comment is accurate, then we have

$I=\displaystyle\int_{3/2}^2\sqrt{(x-1)(3-x)}\,\mathrm dx$

Expand the polynomial, then complete the square within the square root:

$(x-1)(3-x)=-x^2+4x-3=1-(x-2)^2$

$I=\displaystyle\int_{3/2}^2\sqrt{1-(x-2)^2}\,\mathrm dx$

Let $x=2-\cos\theta$ and $\mathrm dx=\sin\theta\,\mathrm d\theta$:

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-(2-\cos\theta-2)^2}\sin\theta\,\mathrm d\theta$

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{1-\cos^2\theta}\sin\theta\,\mathrm d\theta$

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sqrt{\sin^2\theta}\sin\theta\,\mathrm d\theta$

Recall that $\sqrt{x^2}=|x|$ for all $x$, but for all $\theta$ in the integration interval we have $\sin\theta>0$. So $\sqrt{\sin^2\theta}=\sin\theta$:

$I=\displaystyle\int_{\pi/3}^{\pi/2}\sin^2\theta\,\mathrm d\theta$

Recall the double angle identity,

$\sin^2\theta=\dfrac{1-\cos(2\theta)}2$

$I=\displaystyle\frac12\int_{\pi/3}^{\pi/2}(1-\cos(2\theta))\,\mathrm d\theta$

$I=\dfrac\theta2-\dfrac{\sin(2\theta)}4\bigg|_{\pi/3}^{\pi/2}$

$I=\dfrac\pi4-\left(\dfrac\pi6-\dfrac{\sqrt3}8\right)=\boxed{\dfrac\pi{12}+\dfrac{\sqrt3}8}$

You can determine the more general result in the same way.

$I=\displaystyle\int_p^q\sqrt{(x-a)(b-x)}\,\mathrm dx$

Complete the square to get

$(x-a)(b-x)=-(x-a)(x-b)=-x^2+(a+b)x-ab=\dfrac{(a+b)^2}4-ab-\left(x-\dfrac{a+b}2\right)^2$

and let $c=\frac{(a+b)^2}4-ab$ for brevity. Note that

$c=\dfrac{(a+b)^2}4-ab=\dfrac{a^2-2ab+b^2}4=\dfrac{(a-b)^2}4$

$I=\displaystyle\int_p^q\sqrt{c-\left(x-\dfrac{a+b}2\right)^2}\,\mathrm dx$

Make the following substitution,

$x=\dfrac{a+b}2-\sqrt c\,\cos\theta$

$\mathrm dx=\sqrt c\,\sin\theta\,\mathrm d\theta$

and the integral reduces like before to

$I=\displaystyle\int_P^Q\sqrt{c-c\cos^2\theta}\,\sin\theta\,\mathrm d\theta$

where

$p=\dfrac{a+b}2-\sqrt c\,\cos P\implies P=\cos^{-1}\dfrac{\frac{a+b}2-p}{\sqrt c}$

$q=\dfrac{a+b}2-\sqrt c\,\cos Q\implies Q=\cos^{-1}\dfrac{\frac{a+b}2-q}{\sqrt c}$

$I=\displaystyle\frac{\sqrt c}2\int_P^Q(1-\cos(2\theta))\,\mathrm d\theta$

(Depending on the interval [p, q] and thus [P, Q], the square root of cosine squared may not always reduce to sine.)

Resolving the integral and replacing c, with

$c=\dfrac{(a-b)^2}4\implies\sqrt c=\dfrac{|a-b|}2=\dfrac{b-a}2$

because $a$, gives

$I=\dfrac{b-a}2(\cos(2P)-\cos(2Q)-(P-Q))$

Without knowing p and q explicitly, there's not much more to say.