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.
 A: Alternate solution
$$\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cos(x)} dx=\int \frac{\tan(x)}{2\cos^2(x)+1} dx= \int \frac{1}{\cos^2(x)} \frac{\tan(x)}{2+\sec^2(x)} dx$$
Thus, after $t= \tan(x)$ you get
$$\int  \frac{t dt}{t^2+3} $$
A: From the last integral, use $\frac{1}{t(2t^2+1)}=\frac{1}{t}-\frac{2t}{2t^2+1}$. Now, you have:
$$\int \frac{1}{t\cdot(2t^2+1)} \, \mathrm{d}t=\int \frac{1}{t} \, \mathrm{d}t-\int \frac{2t}{2t^2+1} \, \mathrm{d}t=\ln|t|-\frac{1}{2}\ln|2t^2+1|+C$$
A: $$
\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx = \int \frac{1}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big)
$$
$$
= \int \frac{1}{3\cos^3(x)+(1-\cos^2 x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) = \int \frac{1}{3u^3 + (1-u^2)u}\,(-du)
$$
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.
A: $$
\begin{aligned}
& \int \frac{\sin x}{3 \cos ^{3} x+\sin ^{2} x \cos x} d x \\
=& \int \frac{\sin x d x}{\cos x\left(3 \cos ^{2} x+\sin ^{2} x\right)} \\
=& \int \frac{d(\cos x)}{\cos x\left(2 \cos ^{2} x+1\right)} \\
=& \int\left(\frac{1}{\cos x}-\frac{2 \cos x}{2 \cos ^{2} x+1}\right) d(\cos x) \\
=& \ln (\cos x)-\frac{1}{2} \int \frac{d\left(2\cos ^{2} x+1\right)}{2 \cos ^{2} x+1} \\
=& \ln |\cos x|-\frac{1}{2} \ln \left(2 \cos ^{2} x+1\right)+C
\end{aligned}
$$
