Patterns in indefinite integral (1)

Question

So far, I've found out this:

$\int \arcsin x \mathrm d x = x \arcsin x + \sqrt{1-x^2}+C$
$\int x\arcsin x \mathrm d x = \dfrac{x^2}{2}\arcsin x +\dfrac {x}{4}\sqrt{1-x^2} -\dfrac{1}{4}\arcsin x + C$
$\int x^2 \arcsin x \mathrm d x = \dfrac {x^3}{3}\arcsin x+\dfrac{x^2}{9}\sqrt{1-x^2}+\dfrac{2}{9}\sqrt{1-x^2} + C$
$\int x^3 \arcsin x \mathrm d x = \dfrac{x^4}{4}\arcsin x+\dfrac{x^3}{4\times 4}\sqrt{1-x^2}+\dfrac{3x}{4\times 4\times 2}\sqrt{1-x^2}-\dfrac{3\times 1}{4\times 4\times 2}\arcsin x + C$
$\int x^4 \arcsin x \mathrm d x = \dfrac {x^5}{5}\arcsin x +\dfrac {x^4}{5 \times 5}\sqrt{1-x^2}+\dfrac{4x^2}{5\times 5\times 3}\sqrt{1-x^2} + \dfrac{4\times2}{5\times 5\times 3}\sqrt{1-x^2}+C$
$\int x^5 \arcsin x \mathrm d x = \dfrac {x^6}{6}\arcsin x +\dfrac {x^5}{6\times6}\sqrt{1-x^2} +\dfrac{5x^3}{6\times6\times4}\sqrt{1-x^2} +\dfrac{5\times3x}{6\times6\times4\times2}\sqrt{1-x^2} -\dfrac{5\times3\times1}{6\times6\times4\times2}\arcsin x + C$

There seems to be a pattern. It seems like for any positive whole number $n$, $\int x^n \arcsin x \mathrm d x = \mathrm{(polynomial)}\arcsin x + \mathrm{(polynomial)}\sqrt{1-x^2} + C$
So, what is the pattern? What should the above polynomial be?

Answer 1

Here is what Wolfram alpha thinks about the solution of $$\int x^n \arcsin (x) \;\text d x$$ Now, the hypergeometric function is of course a polynomial. So, your guess for the pattern was quite right.

Now, Wolfram doesn't give you the steps. But, once you have the solution, it's very easy to prove that using induction. Note that in the induction step, you just need to do $$\int x^{n+1} \arcsin (x) \;\text{d}x = \int x\cdot x^n \arcsin (x) \;\text {d}x$$ and integrate by parts.

Scroll down the link I embedded to have the formula of differentiating the hypergeometric series. You will need it in the by parts method.

I hope, that helps.

Answer 2

Note: please help with formatting
Here's what I found:

$\int x^n \arcsin x \mathrm d x$
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \int \dfrac{x^{n+1}}{(n+1)\sqrt{1-x^2}}\mathrm d x$
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \dfrac{1}{n+1}\int \dfrac{x^{n+1}}{\sqrt{1-x^2}}\mathrm d x$
[ let x = sin u ]
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \dfrac{1}{n+1}\int \dfrac{(\sin u)^{n+1}}{\sqrt{1-(\sin u)^2}}\mathrm d (\sin u)$
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \dfrac{1}{n+1}\int (\sin u)^{n+1}\mathrm d u$
[ see integration chart ]
$= \dfrac{x^{n+1}}{n+1}\arcsin x - \dfrac{1}{n+1} \left( -\dfrac{1}{n+1}(\sin u)^n\cos u +\dfrac{n}{n+1}\int (\sin u)^{n-1}\mathrm d u \right)$
$= \cdots$
[ sin u = x ]
$= \dfrac{x^{n+1}}{n+1}\arcsin x + \dfrac{x^n}{(n+1)^2}\sqrt{1-x^2} + \dfrac{n x^{n-2}}{(n+1)^2(n-1)}\sqrt{1-x^2}\\ + \dfrac{n}{(n+1)^2}\cdot \dfrac{(n-2)}{(n-1)}\cdot \dfrac{x^{n-4}}{(n-3)}\sqrt{1-x^2} + \cdots \\ + \dfrac{n}{(n+1)^2}\cdot \dfrac{(n-2)}{(n-1)}\cdot \dfrac{x^{n-4}}{(n-3)}\cdots \dfrac{5}{6}\cdot \dfrac{3}{4}\cdot \dfrac{x^1}{2}\sqrt{1-x^2} - \dfrac{n}{(n+1)^2}\cdot \dfrac{(n-2)}{(n-1)}\cdot \dfrac{x^{n-4}}{(n-3)}\cdots \dfrac{5}{6}\cdot \dfrac{3}{4}\cdot \dfrac{1}{2}\arcsin x + C$ [... when n is odd]

Or
$\dfrac{x^{n+1}}{n+1}\arcsin x + \dfrac{x^n}{(n+1)^2}\sqrt{1-x^2} + \dfrac{n x^{n-2}}{(n+1)^2(n-1)}\sqrt{1-x^2}\\ + \dfrac{n}{(n+1)^2}\cdot \dfrac{(n-2)}{(n-1)}\cdot \dfrac{x^{n-4}}{(n-3)}\sqrt{1-x^2} + \cdots \\ + \dfrac{n}{(n+1)^2}\cdot \dfrac{(n-2)}{(n-1)}\cdot \dfrac{x^{n-4}}{(n-3)}\cdots \dfrac{4}{5}\cdot \dfrac{2}{3}\cdot \sqrt {1-x^2} + C$ [...when n is even]