How can I evaluate this integral?
$$\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx$$
2026-04-29 22:41:26.1777502486
Indefinite Integral $\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx$
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Write $$3\sin x+2\cos x=A(2\sin x+3\cos x)+B\frac{d(2\sin x+3\cos x)}{dx}$$
$$\implies 3\sin x+2\cos x=A(2\sin x+3\cos x)+B(2\cos x-3\sin x)$$ $$\implies 3\sin x+2\cos x=(2A-3B)\sin x+(3A+2B)\cos x$$
Solve for $A,B$ equating the coefficients of $\cos x,\sin x$
So, $$\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx=A+B\int\frac{d(2\sin x+3\cos x)}{2\sin x+3\cos x}$$