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I have a problem with this integral

$$\int_\ \frac{\sin 2x }{ \sqrt{4-\cos^2 x}} \, dx$$

We can transform it to

$$\int_\ \frac{2\sin x \cos x }{ \sqrt{4-\cos^2 x}} \, dx$$

Using substitution $u^2 = 4 - \cos^2 x $ we get

$$\int_\ \frac{2u }{\ u } \, du$$

And it gives bad result. Can you point when did i make the mistake ?

davoid
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    Apart from a missing $2$, it looks good to me. We have $2u,du=2\sin x\cos x,dx$. So the integral is $2u+C$. – André Nicolas Jan 30 '16 at 20:45
  • Yeah I forgot about this 2, but result from wolfram is very different. I added this 2. – davoid Jan 30 '16 at 20:48
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    The answer $2\sqrt{4-\cos^2(x)}+C$ is indeed correct! – Mark Viola Jan 30 '16 at 20:49
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    It might differ from yours by a constant, which would be OK. – André Nicolas Jan 30 '16 at 20:49
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    The derivative of $f(x)=2\sqrt{4-\cos^2x}$ is $f'(x)=2\frac{2\sin x\cos x}{2\sqrt{4-\cos^2x}}=\frac{\sin2x}{\sqrt{4-\cos^2x}}$, so your result is good. – egreg Jan 30 '16 at 20:51
  • Ok, but wolfram gives this as a result http://www.wolframalpha.com/input/?i=integral+sin(2x)%2Fsqrt(4-cos%5E2x). I don't know it why. – davoid Jan 30 '16 at 20:54
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    Why should one care about Alpha, a commercial product? Either (i) it gives you in essence the correct answer that you got, or (ii) it is wrong, in which case that's their problem. – André Nicolas Jan 30 '16 at 22:40

1 Answers1

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The result shown by WolframAlpha is $$ \int\frac{\sin 2x}{\sqrt{4-\cos^2 x}} \, dx = \sqrt{2}(\sqrt{7-\cos2x}-\sqrt{7})+\text{constant} $$ that can be rewritten, discarding additive constants, $$ \sqrt{14-2\cos2x}=\sqrt{14-4\cos^2x+2}= \sqrt{4(4-\cos^2x)}=2\sqrt{4-\cos^2x} $$

This means you made no mistake at all.

egreg
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    @MichaelHardy I think WA is substituting $\cos^2x$ with $\cos2x$ and then doing another simple integral. The method used by the OP is sound. – egreg Jan 30 '16 at 21:43