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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.

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    $\begingroup$ Try $s=t^2$. (And please add $dx$ at the three places where it belongs.) $\endgroup$ – Did Jun 13 '12 at 17:54
  • $\begingroup$ That is, after the substitution $t=\cos(x)$ that was mentioned in a previous version of the post. $\endgroup$ – Did Jun 13 '12 at 17:55
  • $\begingroup$ 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". $\endgroup$ – Michael Hardy Jun 13 '12 at 21:09
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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

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

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  • $\begingroup$ I didn't think to divide for $$cos^2(x)$$ and put $$sec^2(x) = 1+tan^2(x)$$ Thank you for the reply! $\endgroup$ – HelloEveryone Jun 13 '12 at 18:11
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$$ \int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx = \int \frac{1}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) $$ $$ = \int \frac{1}{3\cos^3(x)+(1-\cos^2 x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) = \int \frac{1}{3u^3 + (1-u^2)u}\,(-du) $$ 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.

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  • $\begingroup$ I don't think it's possible to use partial fractions at that point. $\endgroup$ – Gigili Jun 13 '12 at 19:50
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    $\begingroup$ @Gigili : Fortunately you're mistaken; it's quite simple. I've added it to my answer. $\endgroup$ – Michael Hardy Jun 13 '12 at 21:04
  • $\begingroup$ Done. It's two logarithms plus an arctangent. $\endgroup$ – Michael Hardy Jun 13 '12 at 21:07
  • $\begingroup$ Well, now that you simplified it ... $\endgroup$ – Gigili Jun 13 '12 at 21:09
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From the last integral, use $\frac{1}{t(2t^2+1)}=\frac{1}{t}-\frac{2t}{2t^2+1}$. Now, you have: $$\int \frac{1}{t\cdot(2t^2+1)} \, \mathrm{d}t=\int \frac{1}{t} \, \mathrm{d}t-\int \frac{2t}{2t^2+1} \, \mathrm{d}t=\ln|t|-\frac{1}{2}\ln|2t^2+1|+C$$

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

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