$\int \cos^{-1} x \; dx$; trying to salvage an unsuccessful attempt

Question

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

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?

Answer 1

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.

Answer 2

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*}

Answer 3

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.

Answer 4

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.