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I have been scratching my head about this problem all day. If one takes the derivative of $\cos^{-1}\frac{x}{a}$, it is quite straightforward to show that the result is $\frac{-1}{\sqrt{a^2 - x^2}}$. However, when I compute the integral of this result using substitution, I obtain the following:

\begin{align}I = \int \frac{-\text{d} x}{\sqrt{a^2 - x^2}} &= \int \frac{-\text{d} x}{a\sqrt{1 - \frac{x^2}{a^2}}} \end{align} and then setting $\sin u = \frac{x}{a}$, \begin{align} \text{d} (\sin u ) &= \cos u \, \text{d} u = \frac{1}{a}\text{d} x; \\ \Rightarrow I &= \int \frac{-\cos u }{\sqrt{1 - \sin^2 u}}\text{d} u \\ &= \int \frac{-\cos u}{\sqrt{\cos^2 u}}\text{d} u \\ &= \int -\text{d} u = -u. \end{align}

Finally we have \begin{align} \sin u &= \frac{x}{a}, \\ \Rightarrow -u & = I = - \sin^{-1}\frac{x}{a}. \end{align}

Anybody have any insight on this? Would be greatly appreciated. I can't see where I'm going wrong or how these could be equivalent values, the only inkling I have is that the inverse trig ratios are not true functions, but I'm not even sure that this is relevant.

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    $$(1) \;\text{d} (\sin u )\, = \,\cos u \text{ d} u = \frac{1}{a}\text{d} x\iff a\cos (u)\,du= dx $$ (2) You've double-typed $du$. – 2017-02-20

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Hint $-\arcsin,\arccos$ differ by a constant which is $\frac {\pi}{2} $ dont forget the C.

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    Thanks a lot for the response. So the only possible way to ascertain that the constant of integration is in fact $\frac{\pi}{2}$ is an a priori knowledge of the derivative? – 2017-02-20
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    @PinoTrapp Happens all the time, as you can see with the [antiderivative of secant cubed](https://en.wikipedia.org/wiki/Integral_of_secant_cubed). – 2017-02-20
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    @PinoTrapp The constant of integration doesn't necessarily have to be $\pi/2$. It can instead be $\pi/2 + C$. The point being that the two possible answers have the same derivative, so they are both an anti-derivative. – 2017-02-20
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    Thanks to everyone for the help. I had no clarity on this problem at all earlier, got into some very silly pitfalls today – 2017-02-20