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I'm trying to find the $F(x)$ of this function but I don't find how to do it, I need some hints about the solution.

I know that $\sin(2x) = 2\sin(x)\cos(x)$ its help me? It's good way to set $2x$ as $t$? $$\int \frac{1+\sin(2x)}{\operatorname{tg}(2x)}dx$$

EDIT
Its right to do it like that?
$$\int \frac{1+2\sin(x)\cos(x)dx}{\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}}$$ $$\int \left(\frac{\cos^2(x)-\sin^2(x)}{2\sin(x)\cos(x)}+\cos^2(x)-\sin^2(x)\right)dx = \int \left(\operatorname{ctg}(2x)+\frac{1+\cos(2x)}{2}-\frac{1-\cos(2x)}{2}\right)dx$$ and the integral of $\displaystyle \operatorname{ctg}(2x) = \frac{\ln|\sin(2x)|}{2}+C$ and the two other is equal to $\displaystyle \frac{\sin(2x)}{2}+C$
Thanks

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3 Answers

up vote 3 down vote accepted

Effective hint:

Let $\int R(\sin x,\cos x)dx$ wherein $R$ is a rational function respect to $\sin x$ and $\cos x$. If $$R(-\sin x, -\cos x)\equiv R(\sin x, \cos x) $$ then $t=\tan x, t=\cot x$ is a good substitution.

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Effective indeed! +1 –  amWhy Jun 2 '13 at 0:07
    
its $tan(x)$ because the original substitution is $tan(\frac{x}{2})$? –  Ofir Attia Jun 2 '13 at 5:45
    
@OfirAttia: Maybe. But this substitution works fast. –  B. S. Jun 2 '13 at 10:02
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Hint: Set $y={\rm tg} x$ and express all the functions through $y$.

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+1 for my friend and Prof. Boris :-) –  B. S. Jun 2 '13 at 10:03
    
@Babak S. Thank you, my friend! –  Boris Novikov Jun 2 '13 at 13:07
    
I will be so thankful if you have time and take a look at this. If you have any suggestion, please let me know them. Thanks a bunch. :-) –  B. S. Jun 2 '13 at 13:42
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HINT: $$\tan(x)= \dfrac {\sin x}{\cos x}$$

$$I=\int \cot 2x . dx +\int \cos 2x.dx $$

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