Question

integrate: (2cosx-3sinx)/(6cosx+4sinx)

Answer 1

$\frac{1}{2} \ln|3\cos x + 2\sin x| + C$

Answer 2

The given integral is $I = \int \frac{2\cos x - 3\sin x}{6\cos x + 4\sin x} dx$.

1. Observe the denominator: $6\cos x + 4\sin x = 2(3\cos x + 2\sin x)$.
2. Let $u = 3\cos x + 2\sin x$.
3. Calculate the differential $du$: $du = \frac{d}{dx}(3\cos x + 2\sin x) dx$ $du = (-3\sin x + 2\cos x) dx$ $du = (2\cos x - 3\sin x) dx$.
4. Substitute $u$ and $du$ into the integral: The numerator $(2\cos x - 3\sin x) dx$ becomes $du$. The denominator $6\cos x + 4\sin x$ becomes $2u$. So, the integral transforms to: $I = \int \frac{du}{2u}$ $I = \frac{1}{2} \int \frac{1}{u} du$
5. Integrate with respect to $u$: $I = \frac{1}{2} \ln|u| + C$
6. Substitute back $u = 3\cos x + 2\sin x$: $I = \frac{1}{2} \ln|3\cos x + 2\sin x| + C$

Alternatively, noting that $\ln|2(3\cos x + 2\sin x)| = \ln|2| + \ln|3\cos x + 2\sin x|$, the constant $\frac{1}{2}\ln|2|$ can be absorbed into the arbitrary constant $C$. Thus, the result can also be written as: $I = \frac{1}{2} \ln|6\cos x + 4\sin x| + C'$

The most direct and simplified form is: $I = \frac{1}{2} \ln|3\cos x + 2\sin x| + C$