Evaluate $\int_{-1}^1 |x|\arcsin^2x \,\rm{d}x$ 
Evaluate
  $\int_{-1}^1 |x|\arcsin^2x \,\rm{d}x$ 

Now, i have a few questions about this integral . (Note : i've tried to apply to it the following property : $\int_{-1}^1 f(x) \,\rm{d}x$=$2\int_0^1 f(x) \,\rm{d}x$ )
First, this problem looks like another problem which I solved and looked like this: $\int_{-1}^1 |x|\arcsin x \,\rm{d}x$ the only difference being the $\arcsin^2x$(this problem which I solved gives 0 because $\arcsin x$ is an odd function and [-1,1] is symmetrical about $0.$). Why is the problem with $\arcsin x$ different from the one that I presented to you (because the problem with $\arcsin^2x$ actually gives a different answer than $0$)?
Second, I've tried splitting the integral in two (using the property in the beginning)  $2\int_{-1}^0 -x\arcsin^2x\,\rm{d}x  +2\int_0^1 x\arcsin^2x \,\rm{d}x$, but it seems like if I am doing the calculations this way it doesn't work ( the only way it works is by evaluating only $2\int_0^1 x\arcsin^2x \,\rm{d}x $, which actually gives the right answer ). The question is, am I splitting the integrals correctly or not? 
Can someone tell me how to approach those types of problems correctly?
Any help will be appreciated!
 A: Let $f$ be an even function with a primitive on the closed interval $[−\text{n}..\text{n}]$, where $\text{n}>0$.
Then:
$$\int_{-\text{n}}^\text{n}f(x)\space\text{d}x=2\int_0^\text{n}f(x)\space\text{d}x\tag1$$
In your case $f$ is an even function, so we get:
$$\int_{-1}^1\left|x\right|\arcsin^2\left(x\right)\space\text{d}x=2\int_0^1\left|x\right|\arcsin^2\left(x\right)\space\text{d}x=2\int_0^1x\arcsin^2\left(x\right)\space\text{d}x\tag2$$
Substitute $\text{u}=\arcsin\left(x\right)$:
$$2\cdot\frac{1}{2}\int_0^\frac{\pi}{2}\text{u}^2\sin\left(2\text{u}\right)\space\text{du}=\int_0^\frac{\pi}{2}\text{u}^2\sin\left(2\text{u}\right)\space\text{du}\tag3$$
Now, use IBP:
$$\int_0^\frac{\pi}{2}\text{u}^2\sin\left(2\text{u}\right)\space\text{du}=\left[-\frac{\text{u}^2\cos\left(2\text{u}\right)}{2}\right]_0^\frac{\pi}{2}+\int_0^\frac{\pi}{2}\text{u}\cos\left(2\text{u}\right)\space\text{du}\tag4$$
Now, use IBP again:
$$\left[-\frac{\text{u}^2\cos\left(2\text{u}\right)}{2}\right]_0^\frac{\pi}{2}+\int_0^\frac{\pi}{2}\text{u}\cos\left(2\text{u}\right)\space\text{du}=$$
$$\left[-\frac{\text{u}^2\cos\left(2\text{u}\right)}{2}\right]_0^\frac{\pi}{2}+\left[\frac{\text{u}\sin\left(2\text{u}\right)}{2}\right]_0^\frac{\pi}{2}-\frac{1}{4}\int_0^\frac{\pi}{2}\sin\left(\text{u}\right)\space\text{du}\tag5$$
