Integrate the function: 2 x - 3 x 6 x + 4 x - Vidyadip

Question

Integrate the function: $\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}$

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Answer

Let $I = \int {\frac{{2\cos x - 3\sin x}}{{6\cos x + 4\sin x}}dx} $

$ = \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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