Could anyone help me how to solve this indefinite integral?

$$\int{\mathrm dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$$

  • 1
    $\begingroup$ Any ideas on your own? Where did you come up with it? Have you solved something similar? Do you expect there to be a in some sense nice solution? $\endgroup$
    – mickep
    Sep 6, 2015 at 18:19
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    $\begingroup$ Even for $\alpha = 0$ the solution seems to be not so pretty: wolframalpha.com/input/?i=int+1%2Fsqrt(sin%5E3+x%2Bsin+(x)) $\endgroup$
    – Alex Silva
    Jan 10, 2017 at 16:34
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    $\begingroup$ @AlexSilva this is REALLY ugly ^^ $\endgroup$
    – tired
    Jan 10, 2017 at 16:55
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    $\begingroup$ @AlexSilva. Using another CAS, it "look" nicer $$-\frac{\sqrt{2} \cot (x) \sqrt{\cot ^2(x)+2} F_1\left(\frac{1}{2};\frac{1}{4},\frac{1}{2};\frac{3}{2};-\cot ^2(x),-\frac{1}{2} \cot ^2(x)\right)}{\sqrt{7 \sin (x)-\sin (3 x)} \csc ^2(x)^{3/4}}$$ where appears the Appell hypergeometric function of two variables. $\endgroup$ May 22, 2018 at 8:11

1 Answer 1


Firstly let us find the anti-derivative in case $\alpha=0$. This is pretty straightforward: \begin{eqnarray} \int\frac{dx}{\sqrt{\sin(x)^3+\sin(x)}}&\underbrace{=}_{y=\sin(x)}&\int\frac{dy}{\sqrt{1-y^2} \sqrt{y} \sqrt{1+y^2}}\\ &\underbrace{=}_{z=y^4}& \frac{1}{4} \int z^{-7/8} (1-z)^{-1/2} dz\\ &=& \frac{1}{4} B_{[\sin(x)]^4}(\frac{1}{8},\frac{1}{2}) \end{eqnarray} where $B_{.}(,)$ is the incomplete beta function.

Now let us compute the derivative at $\alpha=0$. We have: \begin{eqnarray} -\frac{1}{2} \int \frac{\cos(x)}{(\sin(x)^3+\sin(x))^{3/2}}dx&\underbrace{=}_{y=\sin(x)}&-\frac{1}{2} \int\frac{d y}{(y^3+y)^{3/2}}\\ &\underbrace{=}_{u=-y^2}&\frac{(-1)^{3/4}}{4} B_{-\sin(x)^2}(-\frac{1}{4},-\frac{1}{2}) \end{eqnarray}

Therefore we have: \begin{eqnarray} \int \frac{dx}{\sqrt{\sin(x)^3+\sin(x+\alpha)}} =\frac{1}{4} B_{[\sin(x)]^4}(\frac{1}{8},\frac{1}{2})+ \alpha \frac{(-1)^{3/4}}{4} B_{-\sin(x)^2}(-\frac{1}{4},-\frac{1}{2})+O(\alpha^2) \end{eqnarray}


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