How can I find the area between these two curves? $$y=\sin^{-1}(x) \ \text{and} \ y=\frac{\pi}{2}(x^3)$$
I have encountered two problems while solving this question:

*

*The bounds are very small and in terms of decimals so it very difficult to compute without a calculator

*How can I compute the integral of $\arcsin(x)$?

Any help would be appreciated!
 A: 

*Notice that $(x\arcsin x)'=\arcsin x+\frac{x}{\sqrt{1-x^2}}$ so we have $\int\arcsin x dx=x\arcsin x -\int\frac{x}{\sqrt{1-x^2}}=x\arcsin x+\sqrt{1-x^2}+c.$ Or alternatively, you can use the integration by parts rule of blackpenredpen.


*

*The area question is one of the most asked questions in mathematics. To solve this question better, you first need to understand the region whose area is to be computed.

The domain of $y=\arcsin x$ is $[-1,1]$, so we need to plot the graphs of both functions over this closed interval.
The intersection points are found by the equation: $\arcsin x=\frac{\pi}{2}x^3$. I saw these: $(0,0),(1,\frac{\pi}{2})$ and $(-1,-\frac{\pi}{2})$. But, surprise! There are two more. I missed these: https://www.wolframalpha.com/input?i=arcsinx%3Dpi+x%5E3%2F2 They are $\approx(\pm 0.8835,*)$.
So, since there is a symmetry about the origin,
$$A\approx 2\left(\int_0^{0.8835}(\arcsin x-\frac{\pi}{2}x^3)dx+\int_{0.8835}^{1}(\frac{\pi}{2}x^3-\arcsin x)dx\right).$$
By using 1. we can compute the area integral. But we still need calculators.
A: As @Bob Dodds has already answered your questions, I would like to make a small point using symmetry with respect to the line of equation $y=x$.
$\forall x\in [-1,1],$ $$y=\arcsin x \iff x=\sin y \text{ and } y=\frac{\pi}{2}x^3 \iff x=(\frac{2}{\pi}y)^\frac13$$
Moreover, recall that $$\forall \alpha\neq -1,\int x^{\alpha}dx=\frac{1}{\alpha+1}x^{\alpha+1}+C$$
Then, I think you can continue on this path if you wish.

