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$$ \begin{align} \int \cos^{-1} x \; dx &= \int \cos^{-1} x \times 1 \; dx \end{align} $$

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?

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I'd go with "doomed to fail". Also, you seem to have dropped the $x\arccos x$ term in your last equation. – joriki Dec 6 '11 at 15:44
@joriki Thanks for spotting that omission. I have edited accordingly. And yeah, I may just have to concede - it seems to me that any subsequent IBP applications would be futile. I can't simplify the resulting expression from that point on... – ptrcao Dec 6 '11 at 15:51
Why are you overlooking the obvious substitution after the first application of IBP? – Quixotic Dec 6 '11 at 16:04
@MaX It seems obvious in hindsight, and obvious to practiced mathematicians such as yourself no doubt, but I still have trouble recognising when integration by substitution is appropriate - so I had hoped that in such situations, I was still able - eventually - to arrive at the solution simply by reapplying IBP... – ptrcao Dec 6 '11 at 16:11
I've written up the obvious substitution in an answer below. – Michael Hardy Dec 6 '11 at 18:00

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.

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Thanks for slowing down to explain it - I think I needed it. :) – ptrcao Dec 6 '11 at 18:44

Put $x = \cos{\nu}$. Then you have $dx = -\sin\nu \ d\nu$. So your integral is now $$I =\int \nu \cdot \sin\nu \ d\nu$$

This is easy to evaluate by parts. Take $u = \nu$ and $dv = \sin\nu \ d\nu$. Then you have \begin{align*} I &= uv - \int v \ du \\ &= \Bigl[ -\nu \cdot \cos\nu \Bigr] + \int \cos\nu \ d\nu \end{align*}

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You might have created more work for yourself with the second integration by parts, but have you tried using the substitution you mentioned in the previous step? Assuming your work is correct to that point, the substitution $u=1-x^2$ does appear to lead to something more manageable, even after the second integration by parts.

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Yes, I know integration by substitution works after the first IBP. I haven't tried it after the second IBP. I guess that is one way to salvage it, perhaps you (or someone else) could do the honours of showing me how to productively apply integration by substitution after the 2nd IBP? – ptrcao Dec 6 '11 at 16:35
I had intended to leave that to you as it was marked homework, but it looks like Mark has provided what you have requested. – Mike Dec 6 '11 at 18:52
Yes, it appears so; thank you Mike. Don't worry in any case, the homework isn't graded and I always attempt the question before checking the answer because I'm a good student. – ptrcao Dec 7 '11 at 10:20

So, it looks like you are just having problems with $- \int \frac{x^2}{2} \cdot -x(1-x^2)^{-\frac{3}{2}} \; dx =\frac{1}{2}\int \frac{x^3}{(1-x^2)^{\frac{3}{2}}}dx $.

Lets look at $\int \frac{x^3}{(1-x^2)^\frac{3}{2}}dx$

$$ u=1-x^2 , \text{ then } $$ $$ du=-2xdx \text{ and } $$

$$ u-1=x^2 $$ Rewriting, $$ \begin{array}{l l} \int \frac{x^3}{(1-x^2)^\frac{3}{2}}dx &=-\frac{1}{2}\int \frac{x^2(-2xdx)}{(1-x^2)^\frac{3}{2}}\\ \\ \end{array} $$

So, $$ \begin{array}{l l} -\frac{1}{2}\int \frac{1-u}{u^\frac{3}{2}}du &=-\frac{1}{2}\int (u^{-\frac{3}{2}}-u^{-\frac{1}{2}})du\\ &= \frac{1}{\sqrt{u}}+\sqrt{u}+C\\ \end{array} $$ or, going back to x

$$ \frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}+C $$

getting a common denominator gives the equivalent $$ \frac{2-x^2}{\sqrt{1-x^2}}+C $$ I hope that helps. When you first start learning it is tough to know which method to use! Practice is really one of the best ways to get better at integration problems. I didn't try it, but I bet you could solve this integral by trig-sub too.

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Also, don't worry that some people are saying you missed the obvious solution. You definitely missed the most direct approach to the problem - but that isn't a big deal. For one, struggling with the problem will probably make you more likely to check to see if u-sub will work the next time. Plus you got extra practice. – Mark Dec 6 '11 at 18:20

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