오늘의 계산 45 - An Arc Cosine Integral

Today's integral is not hard to solve, but I think it is worth it relish the symmetry innate in it.

Problem. Prove that for $0 < a < b$, \begin{equation} \int_{a}^{b} \arccos \left( \frac{x}{\sqrt{(a+b)x - ab}} \right) \; dx = \frac{\pi}{4} \frac{(b-a)^2}{a+b}. \label{eqn:orig} \end{equation}

Solution. Let \( \alpha = \sqrt{b/a} > 1 \) and \( \beta = \frac{1}{2}(\alpha - \alpha^{-1}) \). Then by the substitution \( x = \sqrt{ab} u \), \eqref{eqn:orig} is equivalent to \begin{equation} \int_{\alpha^{-1}}^{\alpha} \arccos \left( \frac{u}{\sqrt{(\alpha + \alpha^{-1}) u - 1}} \right) \; du = \frac{\pi \beta^2}{\alpha + \alpha^{-1}}. \label{eqn:var_1} \end{equation} Now by the substitution \( v = \sqrt{(\alpha + \alpha^{-1}) u - 1} \), \eqref{eqn:var_1} is equivalent to \begin{equation} \int_{\alpha^{-1}}^{\alpha} 2 v \arccos \left( \frac{v + v^{-1}}{\alpha + \alpha^{-1}} \right) \; dv = \pi \beta^2 . \label{eqn:var_2} \end{equation} Now performing integration by parts, the left hand side of \eqref{eqn:var_2} becomes \begin{equation} \int_{\alpha^{-1}}^{\alpha} \frac{v^2 - 1}{\sqrt{(\alpha + \alpha^{-1})^2 - (v + v^{-1})^2}} \; dv = \int_{\alpha^{-1}}^{\alpha} \frac{v^2 - 1}{\sqrt{(\alpha - \alpha^{-1})^2 - (v - v^{-1})^2}} \; dv . \label{eqn:var_3} \end{equation} Finally, plug \( v = w + \sqrt{1+w^2} \). Then \eqref{eqn:var_3} reduces to \begin{equation} \int_{-\beta}^{\beta} \frac{w\left(w+\sqrt{1+w^2}\right)^2}{\sqrt{\beta^2 - w^2}\sqrt{1+w^2}} \; dw . \label{eqn:var_4} \end{equation} Now expanding the square in the numerator of \eqref{eqn:var_4}, the odd terms cancel out, yielding \begin{equation*} \int_{-\beta}^{\beta} \frac{2w^2}{\sqrt{\beta^2 - w^2}} \; dw . \end{equation*} This integral is easily calculated by the substitution \( w = \beta \sin\theta \), verifying the identity as desired.

References

1. Evaluating \( \int_a^b \arccos\left(x\,/\sqrt{(a+b)x-ab\,}\,\right)\,\mathrm {d}x \) assuming \( 0 < a < b \)