How to solve $\int_{1}^{2} \frac{\sqrt{(x-1)(2-x)}}{x} \, dx$

Question

By using Wolfram-Alpha, this indefinite integral has been solved. However, it was too complicated to calculate the definite integral. Perhaps, this problem can be solved easily with beta function. I would need this easy solution.

The answer is $\frac{\pi}{2}(\sqrt{2}-\sqrt{1})$

I do not understand why the answer to the definite integral is so clear.

It seems strange to use the order that makes the two terms negative in the square root. Not wrong, since it is positive when you multiply them, but making the numerator $$\sqrt{(x-1)(2-x)}$$ makes more sense. — Thomas Andrews, Jan 20 '25 at 06:31
Knowing your reason for expecting a good closed formula would probably give us an idea of where to start — Thomas Andrews, Jan 20 '25 at 06:56
@HariShankar. Very clever substitution which gives for the integrand $$\cos (2 t)+\frac{8}{\cos (2 t)-3}+3$$ — Claude Leibovici, Jan 20 '25 at 09:22
$$\int_1^2\frac{\sqrt{x-1}\sqrt{2-x}}x,dx=\int_0^1\frac{\sqrt{1-x}\sqrt x}{2-x},dx=\int_0^1\frac{\sqrt x\sqrt{1-x}}{1+x},dx$$can be expressed as an infinite sum of beta functions — user170231, Jan 23 '25 at 19:33

score 3 · Accepted Answer · answered Jan 20 '25 at 08:56

Substitute $x=1+\sin^2(t)$ this transforms the integral to $2 \int_0^{\frac{\pi}{2}} \frac{\sin^2(t) \cos^2(t)}{1+\sin^2(t)} \,\mathrm{d}t$, now apply partial fractions to obtain $\int_0^{\frac{\pi}{2}} (\cos(2t)+3-\frac{4 \sec^2(t)}{2 \tan^2(t)+1} ) \, \mathrm{d}t $ which is now elementary to integrate, giving $ (3t+\sin(2t)/2 -2\sqrt{2}\arctan(\sqrt{2} \tan(t)) \bigg|_0^{\frac{\pi}{2}}$ which equals $\frac{\pi}{2}(3-2\sqrt{2})=\frac{\pi}{2}(\sqrt{2}-1)^2$.

How to solve $\int_{1}^{2} \frac{\sqrt{(x-1)(2-x)}}{x} \, dx$

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