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

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

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

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

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 $$x=\sin ^{-1}\left(\sqrt{3} \sqrt{y}\right)$$ $$dx=\frac{\sqrt{3}}{2 \sqrt{y} \sqrt{1-\sqrt{3} y}}$$ $$\sin(x)=\sqrt{3} \sqrt{y}$$ $$\cos(x)=\sqrt{1-\sqrt{3} y}$$ Now, replacing $$I=\frac{1}{2}\int\frac{1}{ \sqrt{1-y^2}}dy=\frac{1}{2} \sin^{-1}(y)=\frac{1}{2} \sin ^{-1}\left(\frac{\sin ^2(x)}{\sqrt{3}}\right)$$

Hint:

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

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

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

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

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