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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.2011-12-06
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    @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...2011-12-06
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    Why are you overlooking the obvious substitution after the first application of IBP?2011-12-06
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    @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...2011-12-06
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    I've written up the obvious substitution in an answer below.2011-12-06
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    @ptrcao:so you are looking for a solution that do not involve substitution at any stage?2011-12-06
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    @MaX Oh, the question itself doesn't stipulate that, I was simply curious to see if I could still solve the problem with reasonable ease in the event that I didn't recognise the relevance of integration by substitution.2011-12-07

4 Answers 4