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I was trying to calculate $$\int \frac{x^3}{\sqrt{1-x^8}}dx$$

Here is what I did:

Let $u=x^4$

Then $du=4x^3dx$

Therefore we get: $\int \frac{\frac14du}{\sqrt{1-u^2}}$

Let $u=\sin \theta$,

Then $du=\cos\theta d\theta$

Therefore we get $\frac14 \int \frac{\cos\theta d\theta}{\sqrt {cos^2\theta}}$

I was expecting to results because $\sqrt {\cos^2\theta}=\pm \cos\theta$

If $\sqrt {\cos^2\theta}=\cos\theta$,

then we have the result $\frac14\sin^{-1}(x^4)+c$,

If $\sqrt {\cos^2\theta}=-\cos\theta$,

then we have the result $-\frac14\sin^{-1}(x^4)+c$.

Differentiating the first one, we get back to the function under the integral sign, but differentiating the second one, what we get is the function under the integral sign with a minus sign.Could somebody tell me where I did it wrong please?

pxc3110
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    @Integrator That's not obvious at all, any reasons for saying that? – pxc3110 Dec 12 '14 at 17:54
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    You get the second result when cos(x) is negative. The range of arcsin is generally the complement of that region. Basically, the derivative of arcsin can be both + and - depending on how we define it. Go through the derivation of the derivative of arcsin to see why this arises – Asvin Dec 12 '14 at 17:54
  • @Integrator I don't see the connection between my question and the one provided by you. – pxc3110 Dec 12 '14 at 17:58
  • @Integrator The question you mention here deals with the definition of $\sqrt x$. – pxc3110 Dec 12 '14 at 17:59

1 Answers1

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You have in effect let $x^4=\sin\theta$. For completeness, one might observe that $\theta$ is chosen between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. In this case we can say a little more, that $0\le \theta\le \frac{\pi}{2}$, since $x^4$ is non-negative.

But that part does not really matter. For $\cos\theta$ is non-negative in the full interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, so $\sqrt{\cos^2\theta}=\cos\theta$.

André Nicolas
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