Integrate the function frac{2 cos x-3 sin x}{ | vedclass

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Question

Let $I = \int \frac{2 \cos x - 3 \sin x}{6 \cos x + 4 \sin x} dx$.We can rewrite the denominator as $2(3 \cos x + 2 \sin x)$.So,$I = \int \frac{2 \cos

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Let $I = \int \frac{2 \cos x - 3 \sin x}{6 \cos x + 4 \sin x} dx$.
We can rewrite the denominator as $2(3 \cos x + 2 \sin x)$.
So,$I = \int \frac{2 \cos x - 3 \sin x}{2(3 \cos x + 2 \sin x)} dx$.
Let $t = 3 \cos x + 2 \sin x$.
Differentiating with respect to $x$,we get $dt = (-3 \sin x + 2 \cos x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{2t} = \frac{1}{2} \int \frac{1}{t} dt$.
Integrating,we get $I = \frac{1}{2} \log |t| + C$.
Substituting back $t = 3 \cos x + 2 \sin x$,we get $I = \frac{1}{2} \log |3 \cos x + 2 \sin x| + C$,where $C$ is an arbitrary constant.

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

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