I was solving the integral $\int\sqrt{2x-x^2}\,\mathrm dx$. I divided it as $\int\sqrt{x}\sqrt{2-x}\,\mathrm dx$ and used the substitution for $x=2\sin^2t$ and $\mathrm dx=4\sin t\cos t\,\mathrm{d}t$

The result I found was $$\arcsin\left(\sqrt{\frac{x}{2}}\right)-\frac{\sqrt{2x-x^2}}{2}({1-x})+C$$

However, on an online calculator I saw the result was $$\frac{(x-1)\sqrt{2x-x^2}+\arcsin(x-1)}{2}+C$$ Apparently it used the substitution with $u=1-x$ respectively $u =\sin t$ and rewriting the integral as $\int \cos^2t\,\mathrm dt$ it found the solution for $t$ and switched the substituted elements.

I understood the solution given there but why my solution gave a different result than that? I had a similar problem in another question as well. And the only difference is with $\arcsin$, I thought it was because I substituted with $2\sin^2x$ which is a power of a trigonometric function but I don't understand why that would be a problem. Is it actually related to that, or is there another reason why it didn't work?


1 Answer 1


The answer is that both solutions are valid: $\frac12\arcsin(x-1)$ and $\arcsin\left(\sqrt{\frac x2}\right)$ differ by a constant.

To see why this is true, suppose that $$\theta = \arcsin\left(\sqrt{\frac x2}\right).$$ Then $$\frac x2 = \sin^2\theta = \frac{1-\cos(2\theta)}2.$$ So $\cos(2\theta) = 1-x$. But $\cos(2\theta) = -\sin(2\theta-\frac\pi2)$, so $$\theta = \frac12\arcsin(x-1)+\frac\pi4.$$

  • $\begingroup$ I understand now. Thank you very much! $\endgroup$
    – MOLD
    Mar 23, 2020 at 16:06

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