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$$ \int\frac{\sin\left(x\right)\cos\left(x\right)}{\sqrt{3 - x^{4}}}{\rm d}x . $$

Hello. Today, in exam, we had to evaluate this integral. Noone was able to do it. Appreciate any help.

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  • $\begingroup$ Try use $x^2=\sqrt{3}\sin \theta$ $\endgroup$ – Mann May 13 '15 at 8:48
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    $\begingroup$ What is the domain of integration ? You're not telling for which values you are integrating. $\endgroup$ – Nihl May 13 '15 at 8:49
  • $\begingroup$ This was indefinite integral... Ok, I´ll try this substitution. $\endgroup$ – Pompo May 13 '15 at 8:55
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    $\begingroup$ This integral doesn't even seem elementary are you sure it's doable? @Pompo $\endgroup$ – Mann May 13 '15 at 9:10
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    $\begingroup$ @Pompo. I should really like to know what answer you will be given for this problem. Please, put it later in your post. $\endgroup$ – Claude Leibovici May 13 '15 at 9:13
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As said in comments, the integral you posted does not seem to be doable.

Let us admit a typo in the problem and let us consider $$I=\int\frac{\sin\left(x\right)\cos\left(x\right)}{\sqrt{3 - \sin^4(x)}}\,{\rm d}x $$ The denominator looking more or less as $\sqrt{1-y^2}$, I thought that $\sin^{-1}(...)$. So, let us try a change of variable such that $$\frac{\sin ^2(x)}{\sqrt{3}}=y$$ that is to say $$x=\sin ^{-1}\left(\sqrt[4]{3} \sqrt{y}\right)$$ $$dx=\frac{\sqrt[4]{3}}{2 \sqrt{y} \sqrt{1-\sqrt{3} y}}$$ $$\sin(x)=\sqrt[4]{3} \sqrt{y}$$ $$\cos(x)=\sqrt{1-\sqrt{3} y}$$ Now, replacing $$I=\frac{1}{2}\int\frac{1}{ \sqrt{1-y^2}}dy=\frac{1}{2} \sin^{-1}(y)=\frac{1}{2} \sin ^{-1}\left(\frac{\sin ^2(x)}{\sqrt{3}}\right)$$

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Hint:

$\int\dfrac{\sin \cos x}{\sqrt{3-x^4}}dx$

$=\int\dfrac{\sin2x}{2\sqrt{3-x^4}}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^nx^{2n+1}}{(2n+1)!\sqrt{3-x^4}}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n-1}x^{2n}}{(2n+1)!\sqrt{3-x^4}}d(x^2)$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n-1}u^n}{(2n+1)!\sqrt{3-u^2}}du$ $(\text{Let}~u=x^2)$

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