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Q. The value of $\int \frac{2 \sin x+3 \cos x}{2 \cos x+3 \sin x} d x$

Integrals

Solution:

$\int \frac{2 \sin x+3 \cos x}{2 \cos x+3 \sin x} d x$
$ \Rightarrow 2 \sin x+3 \cos x=A \frac{d}{d x}(2 \cos x+3 \sin x)+B(2 \cos x+3 \sin x)$
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$\Rightarrow A=\frac{5}{13} B=\frac{12}{13}$
so $\int \frac{2 \sin x+3 \cos x}{2 \cos x+3 \sin x} d x=\frac{5}{13} \int \frac{3 \cos x-2 \sin x}{2 \cos x+3 \sin x} d x+\frac{12}{13} \int 1 \cdot d x$
$\frac{5}{13}$ ln $|2 \cos x+3 \sin x|+\frac{12}{13} x+C$