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Find $\int \frac{1}{\sqrt{1-x^2}}dx$

I can do this using $x = \sin \theta$ and obtain the correct anti-derivative of $\sin^{-1} x + c$. But I wanted to do it using $x = \cos \theta$ since this also simplifies down...

$$\int\frac{1}{\sqrt{1-x^2}}dx = \int \frac{1}{\sqrt{1-\cos^2\theta}} (-\sin\theta) d\theta$$ $$= \int \frac{-\sin \theta}{\sin \theta} d\theta$$ $$ = - \theta + c $$ $$= - \cos^{-1}(x) + c $$

Now how is this the same as $\sin^{-1} x + c$ or how have I gone wrong?

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    $\begingroup$ Their difference is a constant. $\endgroup$ – Botond Aug 19 '18 at 12:13
  • $\begingroup$ Is there a proof for this? $\endgroup$ – PhysicsMathsLove Aug 19 '18 at 12:13
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    $\begingroup$ Yes. The derivative of the difference is zero. $\endgroup$ – Botond Aug 19 '18 at 12:14
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    $\begingroup$ $\arcsin x+\arccos x=\frac\pi2$. $\endgroup$ – Bernard Aug 19 '18 at 12:16
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    $\begingroup$ note: $(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$ and $(-\arccos x)'=-(\arccos x)'=-\left(-\frac{1}{\sqrt{1-x^2}}\right)=\frac{1}{\sqrt{1-x^2}}$. $\endgroup$ – farruhota Aug 19 '18 at 13:51
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Both answers are correct.

The reason is that $$\sin^{-1}x + \cos ^{-1} x $$ is a constant which means $$\sin^{-1}x =-\cos ^{-1} x +C $$

The constant of integration takes care of the apparent difference of the two results.

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