Can't solve integral of Landau Lifschitz Classical Mechanics I am reading right now a book about classical mechanics (Landau Lifschitz) which contains an integral which I do not understand
$\int_{0}^{\alpha}\left [ \frac{dx_{2}(U)}{dU}- \frac{dx_{1}(U)}{dU}\right]dU\int_{U}^{\alpha}\frac{dE}{\sqrt{(\alpha - E)(E-U)}}$
Then the book states the second integral is elementary and results in $\pi$.
And the result is written as:
$\pi\left[ x_{2}(U) - x_{1}(U) \right ]$.
I am not able to solve the second part of the integral nor understand it properly. Can anybody give me a hint on how to start. I was googling already some integral tables but was not able to find it. 
 A: This is a $\int\frac{du}{\sqrt{a^2 - u^2}}$ form, which you can find in integral tables and which most relatively thorough elementary calculus texts discuss. For a very complete treatment of the kinds of classical integration methods that would be familiar to students of the Landau/Lifschitz era, see A Treatise on the Integral Calculus, Volume I, by Joseph Edwards.
To put your integral into this form, complete the square of
$$(\alpha - E)(E-U) = -E^2 + (\alpha + U)E - \alpha U,$$
which gives you
$$ -\left(E \; - \; \frac{\alpha + U}{2} \right)^2 \;\; + \;\; \frac{(\alpha + U)^2}{4} \; - \; \alpha U  $$
Therefore,
$$\;a \; = \; \sqrt{\frac{(\alpha + U)^2}{4} \; - \; \alpha U \;} \;\;\; \text{and} \;\;\; u = E \; - \; \frac{\alpha + U}{2}$$
or
$$\;a \; = \; \sqrt{\frac{(\alpha - U)^2}{4}\;} \;\;\; \text{and} \;\;\; u = E \; - \; \frac{\alpha + U}{2}$$
or
$$\;a \; = \; \frac{1}{2}\alpha \; - \; \frac{1}{2}U \;\;\; \text{and} \;\;\; u = E \; - \; \frac{1}{2}\alpha - \frac{1}{2}U$$
For the variable change $u = E - \frac{1}{2}\alpha - \frac{1}{2}U,$ we have $du = dE$ and the limits $E = U$ to $E = \alpha$ become $u = \frac{1}{2}U - \frac{1}{2}\alpha$ to $u = \frac{1}{2}\alpha - \frac{1}{2}U.$
Thus, we get
$$ \int_{E=U}^{E=\alpha}\frac{dE}{\sqrt{(\alpha - E)(E-U)}} \;\; = \;\; \int_{E=U}^{E=\alpha}\frac{du}{\sqrt{a^2 - u^2}} \;\; = \;\; \int_{u = \frac{1}{2}U - \frac{1}{2}\alpha}^{u = \frac{1}{2}\alpha - \frac{1}{2}U}\frac{du}{\sqrt{a^2 - u^2}}$$
$$= \;\; \left[ \sin^{-1}\left(\frac{u}{a}\right) \right]_{\frac{1}{2}U - \frac{1}{2}\alpha}^{\frac{1}{2}\alpha - \frac{1}{2}U} \;\; = \;\; \sin^{-1}\left(\frac{\frac{1}{2}\alpha - \frac{1}{2}U}{a}\right) \; - \; \sin^{-1}\left(\frac{\frac{1}{2}U - \frac{1}{2}\alpha}{a}\right)  $$
$$ = \;\; \sin^{-1}\left(\frac{\frac{1}{2}\alpha - \frac{1}{2}U}{\frac{1}{2}\alpha - \frac{1}{2}U}\right) \; - \; \sin^{-1}\left(\frac{\frac{1}{2}U - \frac{1}{2}\alpha}{\frac{1}{2}\alpha - \frac{1}{2}U}\right)$$
$$= \;\; \sin^{-1}(1) - \sin^{-1}(-1) \;\; = \;\; \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \;\; = \;\; \pi  $$
