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:

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:

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