<s>[INST] Integral of $\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx$ So, from here $$\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)} dx$$
I divided by cos(x) and I got
$$\int \frac{\tan(x)}{2\cos^2(x)+1} dx$$
But I'm stuck here.
I tried to substitute $t=\cos(x)$
$$\int \frac{-1}{t\cdot(2t^2+1)} dt$$
Any help would be greatly appreciated.
 [/INST] $$
\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx = \int \frac{1}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big)
$$
$$
= \int \frac{1}{3\cos^3(x)+(1-\cos^2 x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) = \int \frac{1}{3u^3 + (1-u^2)u}\,(-du)
$$
Then use partial fractions.
Later edit in response to comments:
$$
\int \frac{1}{3u^3 + (1-u^2)u}\,(-du) = \int\frac{-du}{u(2u^2 + 1)}
 = \int \frac{A}{u} + \frac{Bu+C}{2u^2+1} \, du
$$
Two logarithms plus an arctangent.
</s>