Question
Integrate the function: $\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}$
$ = \int {\frac{{2\cos x - 3\sin x}}{{2\left( {2\sin x + 3\cos x} \right)}}dx} $
$= \frac{1}{2}\int {\frac{{2\cos x - 3\sin x}}{{2\sin x + 3\cos x}}dx} $…(i)
Putting 2 sin x + 3 cos x = t
$ \Rightarrow 2\cos x - 3\sin x = \frac{{dt}}{{dx}}$
$ \Rightarrow $ (2 cos x - 3 sin x)dx = dt
$\therefore$ From eq. (i), $I = \frac{1}{2}\int {\frac{{dt}}{t} = \frac{1}{2}\log \left| t \right| + c} $
$= \frac{1}{2}\log \left| {2\sin x + 3\cos x} \right| + c$
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| $x_i$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
| $p_i$ | $0.1$ | $0.2$ | $0.4$ | $0.2$ | $0.1$ |