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Integrate the function
$\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}$
$\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}$
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Verified Answer
Put $2 \sin x+4 \cos x=t \Rightarrow(2 \cos x-3 \sin x) d x=d t$
$\frac{1}{2} \int \frac{2 \cos x-3 \sin x}{2 \sin x+3 \cos x} d x=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log |t|+c$
$=\frac{1}{2} \log |2 \sin x+3 \cos x|+c$
$\frac{1}{2} \int \frac{2 \cos x-3 \sin x}{2 \sin x+3 \cos x} d x=\frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \log |t|+c$
$=\frac{1}{2} \log |2 \sin x+3 \cos x|+c$
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