Problem source

Using the substitution $x=\alpha\cos^{2}\theta+\beta\sin^{2}\theta,$ show that, if $\alpha<\beta$, 
	\[
	\int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x=\pi.
	\]
What is the value of the above integral if $\alpha>\beta$?
Show also that, if $0<\alpha<\beta$, 
	\[
	\int_{\alpha}^{\beta}\frac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x=\frac{\pi}{\sqrt{\alpha\beta}}.
	\]

Solution source

Using the suggested substitution, we can find.
\begin{align*}
&& x &=\alpha\cos^{2}\theta+\beta\sin^{2}\theta \\
&& x-\alpha &=\alpha(\cos^{2}\theta-1)+\beta\sin^{2}\theta \\
&&& = (\beta - \alpha) \sin^2 \theta \\
&& \beta - x &= -\alpha\cos^{2}\theta+\beta(1-\sin^{2}\theta) \\
&&&= (\beta-\alpha)\cos^2 \theta \\
&& x &=\alpha\cos^{2}\theta+\beta\sin^{2}\theta \\
\Rightarrow && \frac{dx}{d\theta} &= (\beta - \alpha) 2 \cos \theta \sin\theta \\
\\
&&\int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x &= \int_0^{\pi/2} \frac{1}{(\beta - \alpha)\sin\theta \cos \theta} (\beta - \alpha) 2 \cos \theta \sin \theta \, d \theta \\
&&&= \int_0^{\pi/2} \frac{1}{\bcancel{(\beta - \alpha)}\bcancel{\sin\theta \cos \theta}} \bcancel{(\beta - \alpha)} 2 \bcancel{\cos \theta \sin \theta} \, d \theta \\
&&&= \int_0^{\pi/2} 2 d \theta \\
&& &= 2 \frac{\pi}{2} = \boxed{\pi}
\end{align*}

If $\alpha  > \beta$ we can rewrite the integral as:

\begin{align*}
\int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x &= \int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\beta)(\alpha-x)}}\,\mathrm{d}x \\
&= -\int_{\beta}^{\alpha}\frac{1}{\sqrt{(x-\beta)(\alpha-x)}}\,\mathrm{d}x \\
&= -\pi
\end{align*}
Where the last step we are directly using the first integral with the use of $\alpha$ and $\beta$ reversed.

Finally, using the substitution $xt = 1$, we fortunately lose the $\frac1{x}$ term:
\begin{align*}
&& x &= \frac{1}{t} \\
&& \frac{dx}{dt} &= -\frac1{t^2} \\
\\
&& \int_{\alpha}^{\beta}\frac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x &= \int_{\alpha}^{\beta}\frac{t}{\sqrt{(\frac{1}{t}-\alpha)(\beta-\frac{1}{t})}} \frac{-1}{t^2}\,\mathrm{d}t \\
&& &= \int_{\frac1{\alpha}}^{\frac1\beta}\frac{-1}{\sqrt{(1-t\alpha)(t\beta-1)}}\,\mathrm{d}t \\
&& &=  \int_{\frac1{\alpha}}^{\frac1\beta}\frac{-1}{\sqrt{\alpha\beta}\sqrt{(\frac1{\alpha}-t)(t-\frac1{\beta})}}\,\mathrm{d}t \\
&& &=  \frac1{\sqrt{\alpha\beta}}\int_{\frac1{\alpha}}^{\frac1\beta}\frac{-1}{\sqrt{(\frac1{\alpha}-t)(t-\frac1{\beta})}}\,\mathrm{d}t \\
&&&= \boxed{\frac{\pi}{\sqrt{\alpha\beta}}}
\end{align*}
Where again the last step we are using the intermediate integral, with the roles of $\alpha$ and $\beta$ replaced with $\frac{1}{\beta}$ and $\frac1{\alpha}$