Htykuut

Someone the other day told me about the idea of evaluating integrals using horizontal instead of vertical bars (apparently something to do with Lebesque integration but thats way too complicated for me to understand). So I was thinking about this and it occurred to me that and inverse function flips a function that in a way taking its integral is taking the original integral but horizontally. This then lead me to come up with 2 identities I'm not sure are correct and I established graphically.
$$ int_0^xf(x)dx=xf(x)-int_{f(0)}^{f(x)}f^{-1}(x)dx$$
or
$$ int f(x)dx=xf(x)-int f^{-1}(f(x))df(x)$$
although I'm not sure if the second is valid notation but I think I've seen something like that written before somewhere. These are only valid in ranges where $f(x)$ is bijective. For example:
$$begin{align}
int arcsin(x)dx &=xarcsin(x)-int sin(arcsin(x)d(arcsin(x))\
& = xarcsin(x)+cos(arcsin(x)) \
& = xarcsin(x)+sqrt{1-x^2} end{align}$$
is only valid for $-1 le xle 1$ which would be true anyway but would also be true if we evaluated the integral of $sin(x)$ this way giving a range of $-frac{pi}{2}le xle frac{pi}{2}$ where the integral is valid. Having done a bit of reading around the subject of integrals I am sort of surprised why this is never taught as it makes some integrals incredibly easy even though they may already be possible. I just think it would be another tool in the arsenal of mathematicians, that I never see used. $$$$
tl;dr: is the identity valid and if so why is it never taught?

Here's a little proof. Hopefully this helps.
$$I(x):=int_0^x f^{-1}(a)mathrm{d}a$$
Substitution $a=f(b)$:
$$therefore I(x)=int_{f^{-1}(0)}^{f^{-1}(x)}bf'(b)mathrm{d}b$$
Integration by parts:
$$mathrm{d}v=f'(b)mathrm{d}bRightarrow v=f(b)\u=bRightarrow mathrm{d}u=mathrm{d}b$$
$$therefore I(x)=bf(b)bigg|_{f^{-1}(0)}^{f^{-1}(x)}-int_{f^{-1}(0)}^{f^{-1}(x)}f(b)mathrm{d}b$$
$$therefore I(x)=xf^{-1}(x)-int_{f^{-1}(0)}^{f^{-1}(x)}f(b)mathrm{d}b$$
$$therefore int_0^x f^{-1}(t)mathrm{d}t=xf^{-1}(x)-int_{f^{-1}(0)}^{f^{-1}(x)}f(t)mathrm{d}t$$
QED

Example:

$$I=intarcsin x mathrm{d}x$$
setting $f^{-1}(x)=arcsin x$ gives $f(x)=sin x$, and $F(x)=int f(x)mathrm{d}x=-cos x$. We know that
$$I=xf^{-1}(x)-(Fcirc f^{-1})(x)+C$$
Which gives:
$$I=xarcsin x-(-cos(arcsin x))+C$$
$$I=xarcsin x+cosarcsin x+C$$
Note the following:
$$alpha=arcsin x$$
$$therefore sinalpha=x$$
Recall and apply the Pythagorean identity: $cos^2alpha=1-sin^2alpha$
$$therefore cosalpha=sqrt{1-x^2}$$
Plug it in:
$$I=xarcsin x+sqrt{1-x^2}+C$$
You got it!