[Math] Indefinite Integral with “sin” and “cos”: $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $

Question

Indefinite Integral with sin/cos

I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$

Answer 1

It is a pity you did not have a minus-sign in the numerator, since $$ D\ln(3\cos x+2\sin x)=\frac{2\cos x-3\sin x}{3\cos x+2\sin x}, $$ but let us see how we can use this fact anyways.

Let us aim at writing $$ \frac{2\cos x+3\sin x}{3\cos x+2\sin x}=c_1 \frac{2\cos x-3\sin x}{3\cos x+2\sin x}+c_2 \frac{3\cos x+2\sin x}{3\cos x+2\sin x} $$ since both those terms are easy to integrate. This leads us to the linear equations $2=2c_1+3c_2$ and $3=-3c_1+2c_2$. The solution to this system is $c_1=-5/13$ and $c_2=12/13$. Thus $$ \int\frac{2\cos x+3\sin x}{3\cos x+2\sin x}\,dx = -\frac{5}{13}\int \frac{2\cos x-3\sin x}{3\cos x+2\sin x}\,dx +\frac{12}{13}\int \frac{3\cos x+2\sin x}{3\cos x+2\sin x}\,dx. $$ I guess you can take it from here?