Integrate \(\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x\)

This integral appeared in Inside Interesting Integrals by Paul Nahin in the problem set of chapter 3. Using Wolfram Alpha, we get

\begin{equation}
\int\limits_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x = \pi
\label{eq:1}
\tag{1}
\end{equation}

Nahin suggests the following trig substitution, \(x = \cos(2y)\).

While the form of the integrand certainly does suggest that some type of trig substitution will work, let us do it with another method. If we write the integral as

\begin{equation}
\int\limits_{-1}^{1} (1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \mathrm{d} x
\end{equation}

this looks like a beta function. From Higher Transcendental Functions (Bateman Manuscript), Volume 1, Section 1.5.1, equation 10, we see

\begin{equation}
\mathrm{B}(x,y) = 2^{1-x-y} \int\limits_{0}^{1} (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d} t
\label{eq:2}
\tag{2}
\end{equation}

Let us begin with the original integral and the right half of the interval of integration

\begin{equation}
\int\limits_{0}^{1} (1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \mathrm{d} x = \int\limits_{0}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x
\label{eq:3}
\tag{3}
\end{equation}

Now, let us consider

\begin{equation}
\int\limits_{0}^{1} (1+x)^{-\frac{1}{2}}(1-x)^{\frac{1}{2}} \mathrm{d} x = \int\limits_{0}^{1}\sqrt{\frac{1-x}{1+x}} \mathrm{d} x
\label{eq:4}
\tag{4}
\end{equation}

We let \(x=-y\) to obtain

\begin{equation}
-\int\limits_{0}^{-1} \sqrt{\frac{1+y}{1-y}} \mathrm{d} y,
\label{eq:5}
\tag{5}
\end{equation}

which we can rewrite as

\begin{equation}
\int\limits_{-1}^{0}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x
\label{eq:6}
\tag{6}
\end{equation}

Adding the right hand side of equation \eqref{eq:3} and equation \eqref{eq:6} yields our original integral

\begin{equation}
\int\limits_{-1}^{0}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x + \int\limits_{0}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x = \int\limits_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x
\label{eq:7}
\tag{7}
\end{equation}

Likewise, adding the left hand sides of equations \eqref{eq:4} and \eqref{eq:3} yields

\begin{equation}
\int\limits_{-1}^{0}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x + \int\limits_{0}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x =
\int\limits_{0}^{1} (1+x)^{-\frac{1}{2}}(1-x)^{\frac{1}{2}} \mathrm{d} x + \int\limits_{0}^{1} (1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \mathrm{d} x
\end{equation}

If we combine this result into one integral and rearrange the integrand, we see that it is the same as the integral in \eqref{eq:2} with

\begin{equation}
x=\frac{3}{2} \,\, \mathrm{and} \,\, y=\frac{1}{2}
\end{equation}

Putting it all together, we have

\begin{equation}
\int\limits_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x = 2\mathrm{B}\left(\frac{3}{2},\frac{1}{2}\right) = \pi
\end{equation}

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