<s>[INST] $\int \cos^{-1} x \; dx$; trying to salvage an unsuccessful attempt $$
\begin{align}
\int \cos^{-1} x \; dx &= \int \cos^{-1} x \times 1 \; dx
\end{align}
$$
Then, setting 
$$\begin{array}{l l}
u=\cos^{-1} x & v=x \\
u' = -\frac{1}{\sqrt{1-x^2}} & v'=1\\
\end{array}$$
Then by the IBP technique, we have:
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &=\cos^{-1} (x) \cdot x - \int x \cdot -\frac{1}{\sqrt{1-x^2}} \; dx\\
&= x \cos^{-1} (x) - \int -\frac{x}{\sqrt{1-x^2}} \; dx\\
\end{array}$$
Now at this point suppose I have overlooked the possibility of using integration by substitution (setting $u=1-x^2$) to simplify the second integral.  Instead, I attempt to reapply IBP to the second integral $\int -\frac{x}{\sqrt{1-x^2}} \; dx$.
I let
$$\begin{array}{l l}
u= -\frac{1}{\sqrt{1-x^2}} = -(1-x^2)^{-\frac{1}{2}} \qquad & v= \frac{x^2}{2} \\
u' = - \left( -\frac{1}{2} \right) (1-x^2)^{-\frac{3}{2}} \times -2x = -x(1-x^2)^{-\frac{3}{2}} \qquad & v'=x\\
\end{array}$$
Then by IBP again,
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &= x \cos^{-1} (x) - \left( -\frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot -x(1-x^2)^{-\frac{3}{2}} \; dx \right) \\
\end{array}$$
At this stage, I can see no way to proceed.  Can anyone see a reasonable way to salvage this solution, continuing along this line of reasoning?  Or was approaching the second integral by IBP doomed to fail?
 [/INST] The integral
$$
\int \frac{1}{\sqrt{1-x^2}} x\;dx
$$
is BEGGING for a simple substitution.  Don't integrate by parts here.  Instead, do this:
$$
\begin{align}
u & = 1 - x^2 \\  \\ du & = -2x\;dx \\  \\ \frac{-du}{2} & = x\;dx
\end{align}
$$
You get
$$
-\int \frac{1}{2\sqrt{u}} \;du = \sqrt{u}+C = \sqrt{1-x^2}+ C.
$$
The reason why some people are calling this substitution "obvious" is that you have $\big((\text{1st-degree polynomial})\cdot dx\big)$ and elsewhere in the expression you have a 2nd-degree polynomial, and---lo and behold---the derivative of the 2nd-degree polyonomial is the 1st-degree polyonomial, except for the factor of $1/2$, which is a constant.  When you see something multiplied by $dx$ that is the derivative of another expression that's there, then that's what you do.  You should be looking for that.
Later note: Suppose I am asked about
$$
\int \frac{1}{\sqrt{1-x^2}} x\;dx.
$$
If this were in class or the if the questioner were otherwise physically present, I might just say "Here's a hint", and then write
$$
\int \frac{1}{\sqrt{1-x^2}} {\huge(}x\;dx{\huge)}.
$$
When you understand why that's a hint, then you understand finding integrals by means of substitutions.
</s>