<s>[INST] Proving $\int_a^b \frac {x dx}{\sqrt{(x^2-a^2)(b^2-x^2)}} = \frac {\pi}{2} $ Can anybody here help me to prove that 
$$
\int_a^b \frac {x\, \mathrm dx}{\sqrt{(x^2-a^2)(b^2-x^2)}} = \frac {\pi}{2} 
$$
Thanks for your help.
 [/INST] Making following substitution$$x^2=a^2\sin^2t+b^2\cos^2t\iff x \,\mathrm dx=(b^2-a^2)\sin t\cos t \,\mathrm dt$$
We have
$$\sqrt{(x^2-a^2)(b^2-x^2)}=\sqrt{(b^2-a^2)^2\sin^2t\cos^2t}=(b^2-a^2)\sin t\cos t$$
By putting everything together
$$\int_a^b \frac {x  \,\mathrm dx}{\sqrt{(x^2-a^2)(b^2-x^2)}}=\int_0^{\pi/2} \frac{(b^2-a^2)\sin t\cos t \,\mathrm dt}{(b^2-a^2)\sin t\cos t} =\int_0^{\pi/2}  \,\mathrm dt= \dfrac{\pi}{2}$$

$$\large\int_a^b \frac {x  \,\mathrm dx}{\sqrt{(x^2-a^2)(b^2-x^2)}}=\dfrac{\pi}{2}$$

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