# 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.

• Try $s=t^2$. (And please add $dx$ at the three places where it belongs.) – Did Jun 13 '12 at 17:54
• That is, after the substitution $t=\cos(x)$ that was mentioned in a previous version of the post. – Did Jun 13 '12 at 17:55
• I am inclined to think that the solution I posted below is probably the most straightforward one unless you have one that uses some surprising "trick". – Michael Hardy Jun 13 '12 at 21:09

## 3 Answers

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

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

• I didn't think to divide for $$cos^2(x)$$ and put $$sec^2(x) = 1+tan^2(x)$$ Thank you for the reply! – HelloEveryone Jun 13 '12 at 18:11

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

• I don't think it's possible to use partial fractions at that point. – Gigili Jun 13 '12 at 19:50
• @Gigili : Fortunately you're mistaken; it's quite simple. I've added it to my answer. – Michael Hardy Jun 13 '12 at 21:04
• Done. It's two logarithms plus an arctangent. – Michael Hardy Jun 13 '12 at 21:07
• Well, now that you simplified it ... – Gigili Jun 13 '12 at 21:09

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

• Thank you! You reach that point by usinf partial fraction, am I right? – HelloEveryone Jun 13 '12 at 18:09
• @HelloEveryone: Yes, I did. – Dennis Gulko Jun 13 '12 at 18:16