HP 15C and INT(1/√(1-x),0,1)

[Thomas Klemm]

The HP-15C ADVANCED FUNCTIONS HANDBOOK has on page 47 a section about:

Transformation of Variables

Quote:In many problems where the function changes very slowly over most of a very wide interval of integration, a suitable transformation of variables may decrease the time required to calculate the integral.

You can use \(x=1-t^2\) which leads to:
\[ \begin{eqnarray}
dx&=&-2t\cdot dt \\
\sqrt{1-x}&=&t \\
\int_0^1\frac{1}{\sqrt{1-x}}dx&=&\int_1^0\frac{-2t}{t}dt \\
&=&2\int_0^1dt \\
&=&2t\bigg\vert_0^1 \\
&=&2-0=2
\end{eqnarray} \]

Or then use \(x=\cos^2(t)\) which leads to:
\[ \begin{eqnarray} \\
dx&=&-2\cos(t)\sin(t)dt \\
\sqrt{1-x}&=&\sin(t) \\
\int_0^1\frac{1}{\sqrt{1-x}}dx&=&\int_{\frac{\pi}{2}}^0\frac{-2\cos(t)\sin(t)}{\sin(t)}dt \\
&=&2\int_0^{\frac{\pi}{2}}\cos(t)dt \\
&=&2\sin(t)\bigg\vert_0^{\frac{\pi}{2}} \\
&=&2-0=2
\end{eqnarray} \]

In both cases the resulting function can be integrated without problems.
It's just a coincidence that both integrals can also be calculated easily algebraically.

Cheers
Thomas