Integral of $\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx$

Question

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.

Answer 1

From the last integral, use $\frac{1}{t(2t^2+1)}=\frac{1}{t}-\frac{2t}{2t^2+1}$. Now, you have: $$\int \frac{1}{t\cdot(2t^2+1)} \, \mathrm{d}t=\int \frac{1}{t} \, \mathrm{d}t-\int \frac{2t}{2t^2+1} \, \mathrm{d}t=\ln|t|-\frac{1}{2}\ln|2t^2+1|+C$$

Answer 2

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

$$\int \frac{t dt}{t^2+3} $$

Answer 3

$$ \int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx = \int \frac{1}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) $$ $$ = \int \frac{1}{3\cos^3(x)+(1-\cos^2 x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) = \int \frac{1}{3u^3 + (1-u^2)u}\,(-du) $$ 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.

Answer 4

$$ \begin{aligned} & \int \frac{\sin x}{3 \cos ^{3} x+\sin ^{2} x \cos x} d x \\ =& \int \frac{\sin x d x}{\cos x\left(3 \cos ^{2} x+\sin ^{2} x\right)} \\ =& \int \frac{d(\cos x)}{\cos x\left(2 \cos ^{2} x+1\right)} \\ =& \int\left(\frac{1}{\cos x}-\frac{2 \cos x}{2 \cos ^{2} x+1}\right) d(\cos x) \\ =& \ln (\cos x)-\frac{1}{2} \int \frac{d\left(2\cos ^{2} x+1\right)}{2 \cos ^{2} x+1} \\ =& \ln |\cos x|-\frac{1}{2} \ln \left(2 \cos ^{2} x+1\right)+C \end{aligned} $$