024. Integration of Rational Functions

Evaluate

$$\begin{aligned} \int \cfrac{1}{1 + \tan x} dx \end{aligned}$$

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When the numerator can be expressed in terms of the derivative of the denominator, the expression can be rewritten as follows, and then the constants $\alpha$ and $\beta$ can be determined. $$\begin{aligned} \int \cfrac{g(x)}{f(x)} dx = \int \cfrac{\alpha f(x) + \beta f'(x)}{f(x)} dx = \alpha x + \beta \ln \vert f(x) \vert + C \end{aligned}$$

To make use of this, first multiplying both the numerator and the denominator by $\cos x$ gives $$\begin{aligned} \int \cfrac{\cos x}{\sin x + \cos x} dx = \int \cfrac{\alpha \left( \sin x + \cos x \right)}{\sin x + \cos x} + \cfrac{\beta \left( \cos x - \sin x \right)}{\sin x + \cos x} dx \end{aligned}$$

Since $\cos x = (\alpha - \beta) \sin x + (\alpha + \beta) \cos x$, $\alpha - \beta = 0$ and $\alpha + \beta = 1$, so $\alpha = \beta = 1 / 2$. Therefore, $(x + \ln \vert \sin x + \cos x \vert) / 2 + C$.