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Evaluating $\int P(\sin x, \cos x) \text{d}x$

I am stuck with this integral: \begin{equation} \int \frac{dx}{(3+\cos^2x) \cdot \tan x} \end{equation} I tried playing with trigonometrical function and the most promissing variant I managed to produce was this: \begin{equation} \frac{1}{2}\int \frac{dx}{\sin2x} \end{equation}

I am guessing that to integrate this substitution is required but of what... I need some good instruction how to deal with this.

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marked as duplicate by Aryabhata, tomasz, J. M., sdcvvc, William Aug 29 '12 at 21:55

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Another approach is to use $\cos^2 x=1-\sin^2 x$. Write the $\tan x$ as $(\sin x)/(\cos x)$. We end up with $$\int \frac{\cos x}{{}(4-\sin^2 x)\sin x}\,dx.$$ Make the substitution $u=\sin x$. We end up with $$\int \frac{1}{(4-u^2)u}.$$ This is a rational function, use partial fractions. I think that our integrand is $$\frac{1}{8}\frac{1}{2-u}-\frac{1}{8}\frac{1}{2+u}+\frac{1}{4}\frac{1}{u}.$$ Now the integration is easy. We get $$-\frac{1}{8}\log(4-u^2)+\frac{1}{4}\log u +C.$$

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I have used it but missed one variable while rewriting it. Thanks! –  Povylas Mar 1 '12 at 21:13
    
@Povylas: Had, as pretty often, a minus sign problem. Fixed. –  André Nicolas Mar 1 '12 at 21:30
    
Noticed while checking it :) –  Povylas Mar 1 '12 at 21:34
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Integrating $\int\frac{dx}{\sin 2x}$ is fairly simple. Use $u=2x$. Then $du = 2\,dx$. You'll be left with $\frac{1}{4}\int \csc u\, du$. This is $\ln(\sin x)- \ln(\cos x)$.

However, I don't think this is equivalent to your initial integral. Try calling the denominator of your initial integral $(4\cos^2x + 3\sin^2x) \tan x$. Try to see where it goes from there. We can also make it $\int\frac{\cos x}{(4-\sin^2x)\sin x}\,dx$ and use $u=\sin x$ to get $\int \frac{dx}{(4-u^2)u}$. Completely factorizing the denominator using difference of squares we can then use integration by parts.

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Sorry I can`t use csc... I am limited to sin cos tan and ctan only. –  Povylas Mar 1 '12 at 21:06
    
csc is just 1/sin. –  Ali Mar 1 '12 at 21:11
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