Trigonometric rational functions.

The substitution $ t= 2 \arctan x$ replaces every rational function in $ \sin x$ , $ \cos x$ and $ \tan x$ by a rational function of $ t$ , according to the following formulas:

\fbox{
$\sin x = \frac {2t}{1+t^2} \qquad ; \qquad \cos x = \frac {1-t^2}{1+t^2}...
... ; \qquad \tan x = \frac {2t}{1-t^2}
\qquad ; \qquad dx = \frac {2dt}{1+t^2}$
}
After computing the integral of the rational function of the undeterminate $ t$ . don't forget to return to the original undeterminate $ x$ .

Example 8.2.14   Let $ F(x)=\int \frac {1}{1+\sin x - \cos x} dx$ .

We have: $ \frac {1}{1+\sin x - \cos x}= \frac {1}{1+\frac {2t}{1+t^2} - \frac {1-t^2}{1+t^2}}
=\frac {t^2+1}{2t(t+1)}$ .

Thus: $ F(x)=\int \frac {t^2+1}{2t(t+1)} \cdot \frac {2dt}{1+t^2} = \int \frac {dt}{t(t+1)}.$

Working as in 2.5, we have:

$ F(x)=\int \left( \frac 1t - \frac {1}{t+1} \right) dt
= \ln \vert t\vert - \ln \vert t+1\vert+C = \ln \left\vert \frac {t}{t+1} \right\vert +C$ .

Now we return to the original undeterminate:

$ F(x)=\ln \left\vert \frac {\tan \frac x2}{1+ \tan \frac x2} \right\vert +C$ .



Noah Dana-Picard 2007-12-28