How to Prove This Elegant Integral Identity Involving Trigonometric and Square Root Terms

Question

Let $ a $, and $ b $ be real numbers. Show that: \begin{equation} \int_{0}^{\pi} \frac{\left(\sqrt{1 + (a \cos t + b \sin t)^2} + a \cos t + b \sin t\right)^2}{1 + \left(\sqrt{1 + (a \cos t + b \sin t)^2} + a \cos t + b \sin t\right)^2} \, dt = \arccos\left(-\frac{b}{\sqrt{1 + a^2 + b^2}}\right). \nonumber \end{equation}

I have been trying to prove this integral identity but have encountered some challenges. Here is the reasoning I have followed so far:

Trigonometric substitutions: I attempted to simplify the term $ a \cos t + b \sin t $ using the identity $ r \cos(t - \phi) $, where $ r = \sqrt{a^2 + b^2} $ and $ \phi = \arctan(b/a) $. This seemed natural, but it led to complicated expressions in terms of $ t - \phi $, and I couldn't make significant progress.

Change of variable: I considered a change of variable to take advantage of the symmetry of the interval $ [0, \pi] $, but it wasn’t clear how to simplify the square root in the numerator.

Relation to the arccos function: I suspect that the result involves some key property of trigonometric functions and their relation to the arccos function, but I am unsure how to connect this to the integrand.

This problem seems intriguing because it combines definite integrals, trigonometric terms, square roots, and inverse functions, suggesting there is an elegant simplification underlying it. I would appreciate any guidance on how to proceed or if there is a technique I might be overlooking.

Answer 1

Note that $$\frac{\left(\sqrt{1 + (a \cos t + b \sin t)^2} + a \cos t + b \sin t\right)^2}{1 + \left(\sqrt{1 + (a \cos t + b \sin t)^2} + a \cos t + b \sin t\right)^2} \\ =\frac12\bigg(1+\frac{a \cos t + b \sin t}{\sqrt{1 + (a \cos t + b \sin t)^2} } \bigg) \\ $$ Then \begin{align} I=& \ \frac12\int_0^\pi \bigg(1+\frac{a \cos t + b \sin t}{\sqrt{1 + (a \cos t + b \sin t)^2} } \bigg) dt\\ =&\ \frac\pi2 +\frac12\int_0^\pi \frac{d(a \sin t - b \cos t)}{\sqrt{1+a^2+b^2 - (a \sin t - b \cos t) ^2} } \\ = &\ \pi - \cos^{-1}\frac{b}{\sqrt{1 + a^2 + b^2}}= \cos^{-1}\frac{-b}{\sqrt{1 + a^2 + b^2}} \end{align}

Answer 2

Let $x = a\cos t + b \sin t = r \sin(t+c)$ where $r=\sqrt{a^2+b^2}$, $\sin c = a/r$, and $\cos c = b/r$. After some algebraic manipulations, we can simplify the integrand considerably: $$ \frac{(\sqrt{1+x^2}+x)^2}{1+(\sqrt{1+x^2}+x)^2} = \frac12\left(1 + \frac{x}{\sqrt{1+x^2}}\right). $$

The integral of the first term is simply $\frac\pi2$. The integral of the second term is $$ \frac12\int_0^\pi \frac{r\sin(t+c)}{\sqrt{1+r^2 \sin^2(t+c)}}~dt = \frac12\int_{-b}^b \frac{1}{\sqrt{1+r^2 - y^2}}~dy = \int_{0}^b \frac{1}{\sqrt{1+r^2 - y^2}}~dy, $$ where I substituted $y=r\cos(t+c)$. The antiderivative of the integrand is $-\arccos\left(\frac{y}{\sqrt{1+r^2}}\right)$, so the integral is $$ \int_{0}^b \frac{1}{\sqrt{1+r^2 - y^2}}~dy = -\arccos\left(\frac{b}{\sqrt{1+r^2}}\right) + \frac\pi2. $$ Therefore, the full integral evaluates to $$ \pi - \arccos\left(\frac{b}{\sqrt{1+r^2}}\right) = \arccos\left(-\frac{b}{\sqrt{1+r^2}}\right). $$