How do solve $\int\frac{2\sin(x)+3\cos(x)}{3\sin(x)+2\cos(x)}dx$?

Question

How do I solve this integral? Should I use some kind of an integral substitution?

Answer 1

Hint. By setting $$ I=\int\frac{\cos(x)\:dx}{2\cos(x)+3 \sin(x)}\quad J=\int\frac{\sin(x)\:dx}{2\cos(x)+3 \sin(x)} $$ One may observe that

$$\begin{cases} 2 I+3J=\displaystyle\int 1\:dx \\ 3 I-2J=\displaystyle \int\frac{(2\cos(x)+3 \sin(x))'}{2\cos(x)+3 \sin(x)}\:dx \end{cases} $$ Can you take it from here?

Answer 2

If there is no minus sign, multiply the numerator and the denominator by $ sec (x)^{3} $. Then substitute $ u= tan (x) $ and then the integral becomes easy. Hope it helps.

Answer 3

Hint:

Let $$2\sin x+3\cos x=A(3\sin x+2\cos x)+B\cdot\dfrac{d(3\sin x+2\cos x)}{dx}$$