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.

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    $\begingroup$ What did you try? $\endgroup$ – Did Nov 7 '13 at 15:22
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    $\begingroup$ Perhaps there is a certain substitution that is suggested by the form of the integrand? $\endgroup$ – GEdgar Nov 7 '13 at 15:33
  • $\begingroup$ we have tried x^2=a^2*sin^2+b^2*cos^2, but don't know how to do next? $\endgroup$ – user106162 Nov 7 '13 at 15:40
  • $\begingroup$ Are you user "N C"? // Excellent idea, but did you really try it? $\endgroup$ – Did Nov 7 '13 at 15:41
  • $\begingroup$ we're friends. we don'r know why x^2=a^2*sin^2+b^2*cos^2 $\endgroup$ – N C Nov 7 '13 at 15:50

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 $$\int_a^b \frac {x \,\mathrm dx}{\sqrt{(x^2-a^2)(b^2-x^2)}}=\int_0^{\pi/2} \frac{(b^2-a^2)\sin t\cos t \,\mathrm dt}{(b^2-a^2)\sin t\cos t} =\int_0^{\pi/2} \,\mathrm dt= \dfrac{\pi}{2}$$

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


Let $t=\dfrac{x^2-a^2}{b^2-a^2}$ and $dt=\dfrac{2x\,dx}{b^2-a^2}$, then $$\begin{align}\int_a^b \frac {x \,dx}{\sqrt{(x^2-a^2)(b^2-x^2)}}&=\frac{1}{2}\int_0^1 \frac {dt}{\sqrt{t\,(1-t)}}\\& = \frac{1}{2}\int_0^1 t^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}}\,dt \\ & = \frac{1}{2}\text{B}\left( \frac{1}{2}, \frac{1}{2}\right)\\& =\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(1 \right)}\\&=\frac{\pi}{2}\end{align}$$ Alternatively, we can use substitution $t=\sin^2\theta$ and $dt=2\sin\theta\cos\theta\,d\theta$. $$\begin{align}\int_a^b \frac {x \,dx}{\sqrt{(x^2-a^2)(b^2-x^2)}}&=\frac{1}{2}\int_0^1 \frac {dt}{\sqrt{t\,(1-t)}}\\& = \frac{1}{2}\int_0^{\pi/2}\frac{2\sin\theta\cos\theta\,d\theta}{\sqrt{\sin^2\theta\cos^2\theta}}\\&=\frac{\pi}{2}\end{align}$$

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    $\begingroup$ Nice answer (+1) $\endgroup$ – Aditya Hase Dec 4 '14 at 17:06

After substituting $y=x^2$, this integral may be solved by a type-III Euler substitution, of the form:


In my opinion, Euler substitutions offer an elegant alternative to trigonometric substitutions, and they are unfortunately under-taught in basic calculus courses. So when applicable, I like to post solutions using the Euler method for the sake of raising awareness, even for problems such as this one that have already been adequately solved by other methods. Using the substitution relation $(1)$ above, we find,

$$\begin{cases} t=\frac{\sqrt{\left(y-a^2\right)\left(b^2-y\right)}}{y-a^2},\\ y=\frac{a^2t^2+b^2}{t^2+1},\\ \mathrm{d}y=\frac{2t\left(a^2-b^2\right)}{\left(t^2+1\right)^2}\,\mathrm{d}t,\\ \end{cases}$$

and so the integral evaluates to:

$$\begin{align} I &=\int_{a}^{b}\frac{x\,\mathrm{d}x}{\sqrt{\left(x^2-a^2\right)\left(b^2-x^2\right)}}\\ &=\frac12\int_{a^2}^{b^2}\frac{\mathrm{d}y}{\sqrt{\left(y-a^2\right)\left(b^2-y\right)}}\\ &=\frac12\int_{\infty}^{0}\frac{t^2+1}{|b^2-a^2||t|}\cdot\frac{2t\left(a^2-b^2\right)}{\left(t^2+1\right)^2}\,\mathrm{d}t\\ &=\frac{b^2-a^2}{|b^2-a^2|}\int_{0}^{\infty}\frac{\mathrm{d}t}{t^2+1}\\ &=\operatorname{sgn}{\left(b^2-a^2\right)}\cdot\frac{\pi}{2}.\\ \end{align}$$


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