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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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    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".2012-06-13

3 Answers 3

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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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    I didn't think to divide for $cos^2(x)$ and put $sec^2(x) = 1+tan^2(x)$ Thank you for the reply!2012-06-13
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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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    @HelloEveryone: Yes, I did.2012-06-13
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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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    Well, now that you simplified it ...2012-06-13