Cool integral

$\begin{aligned} I & = \int_a^b \frac {dx}{\sqrt{(x-a)(b-x)}} \\ & = \int_a^b \frac {dx}{\sqrt{\left(\left(x-\frac {a+b}2\right)+\frac {b-a}2\right)\left(\frac {b-a}2-\left(x-\frac {a+b}2\right)\right)}} \\ & = \int_a^b \frac {dx}{\sqrt{\left(\frac {b-a}2\right)^2-\left(x-\frac {a+b}2\right)^2}} \\ & = \int_a^b \frac {dx}{\frac {b-a}2 \sqrt{1-\left(\color{#3D99F6}{\frac {2x-a-b}{b-a}} \right)^2}} \end{aligned}$

Now, let $\sin \theta = \color{#3D99F6}{\dfrac {2x-a-b}{b-a}}$ $\implies \cos \theta \ d \theta = \dfrac 2{b-a} dx$ . When $x=a$ , $\sin \theta = -1$ $\implies \theta = -\frac \pi 2$ . When $x=b$ , $\sin \theta = 1$ $\implies \theta = \frac \pi 2$ .

Then, we have:

$\displaystyle I = \int_{-\frac \pi 2}^\frac \pi 2 \frac {\cos \theta}{\sqrt{1-\sin^2 \theta}} d \theta = \int_{-\frac \pi 2}^\frac \pi 2 \frac {\cos \theta}{\cos \theta} d \theta = \int_{-\frac \pi 2}^\frac \pi 2 d \theta = \theta \ \bigg|_{-\frac \pi 2}^\frac \pi 2 = \boxed{\pi}$ for any real $a$ and $b$ .

My solution-

$\displaystyle\int_1^3 \dfrac{dx}{\sqrt{(x-1)(3-x)}}$

$\displaystyle\int_1^3 \dfrac{dx}{\sqrt{-x^2+4x-3}}$

$\displaystyle\int_1^3 \dfrac{dx}{\sqrt{1-(x-2)^2}}$

Note that $\frac{d[x-2]}{dx}=1$ ,

$\arcsin(x-2)|_1^{3}$

$\arcsin(3-2)-\arcsin(1-2)$

$\arcsin(1)-\arcsin(-1)=\pi≈3.1415$

Notice that, $(x-1)(3-x)$ can be simplified by completing its square, see:

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

Now, take $u=x-2$ , then $du=dx$ .

Following to this, when $x=3,u=1$ and when $x=1,u=-1$ . Hence the integral becomes:

$\displaystyle \int_{-1}^{1} \frac{1}{\sqrt{1-u^2}} du$

Next in house is a trigonometric substitution, take $u=\sin t$ then $du = \cos t dt$

Subsequently, when $u=1,t=\frac{\pi}{2}$ and when $u=-1,t=-\frac{\pi}{2}$ , therefore the integral becomes:

$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} dt = \pi$

Substitute x=acos^2q+bsin^2q. The integral will easily come in terms of dq only. To find the value of q in terms of x use the relations 1+cos2x=2cos^2x and 1-cos2x=2sin^2x q then turns out to be (1/2)arccos((2x-a-b)/(a-b)). Substitute the limits.