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Finding value of $\displaystyle \int \sqrt{x+\sqrt{kx-x^2}}dx$

My Try: Let $\displaystyle I = \sqrt{x+\sqrt{kx-x^2}}$. Substitute $kx-x^2=t^2$

Then $(k-2x)dx = dt$

$\Rightarrow \displaystyle I = \int \sqrt{x+t}\; \times \frac{1}{k-2x}dt$

Now $\displaystyle x^2-kx+t^2=0\Rightarrow x=\frac{k\pm \sqrt{k^2-4t^2}}{2}$

$\displaystyle \Rightarrow I = \pm \int \sqrt{\frac{k\pm \sqrt{k^2-4t^2}}{2}+t}\times \frac{1}{\sqrt{k^2-4t^2}}dt$

Could some help me how to solve it

please help me , Thanks

DXT
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    Use $$x-\dfrac{k}{2}=\dfrac{k}{2}\sin\theta$$ – Nosrati Sep 10 '18 at 08:46
  • It seems that it would end to an Elliptic integral! – Mikasa Sep 10 '18 at 08:50
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    A few steps: With Nosrati's hint $x=\dfrac{k}{2}+\dfrac{k}{2}\sin\theta$:

    $$ I = \int \sqrt{x+\sqrt{kx-x^2}}dx = \ \left[\dfrac{k}{2}\right]^{3/2} \int \sqrt{1+\sin\theta +\cos\theta} ; \cos\theta ; d\theta \ = \left[\dfrac{k}{2}\right]^{3/2} \int \sqrt{\sqrt{2} \sin(\theta + \pi/4) + 1} ; \cos\theta ; d\theta \ $$ .....

     – Andreas Sep 10 '18 at 12:21
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    ... Since $$\cos\theta = \cos(\theta + \pi/4 - \pi/4) = \cos(\theta + \pi/4) \cos(\pi/4) -\sin(\theta + \pi/4) \sin(\pi/4) \= \frac{1}{\sqrt 2}\cos(\theta + \pi/4) - \frac{1}{\sqrt 2}\sin(\theta + \pi/4) $$
    set $\alpha = \theta + \pi/4$ and obtain

    $$ I = \left[\dfrac{k}{2}\right]^{3/2} \int \sqrt{\sqrt{2} \sin(\alpha) + 1} ; (\frac{1}{\sqrt 2}\cos(\alpha) - \frac{1}{\sqrt 2}\sin(\alpha) ) ; d\alpha\ $$ The first one is easy, the second one becomes elliptic

     – Andreas Sep 10 '18 at 12:21

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