Are you allowed to do this (integral)

Question

I have no clue how to explain this in words, but I thought of the integral $$\int_{z=0}^1\int_{\sin x=0}^y\cos x\ d\left(\sin x\right)dz$$ and I'm wondering if it would be possible. I wrote it that way to make it less ambiguous and look less like $\int_0^y\cos x\cdot\frac{d}{dx}\sin x$ as d(sinx) could be ambiguous. I think I've solved it but it could just be meaningless

Edit: For all of you thinking the d(sinx) is a typo it isnt, i want to integrate from sinx=0 to sinx = y, i added the 2nd integral to make it less ambiguous but people ended up thinking its a typo

Answer 1

It is a result of Leibniz that $\frac{dy}{dx}dx=dy$. The chain rule of differentiation and the u-substitution of integration are both essentially the same fact as this. What you did was simply a u-substitution, and perfectly valid.

Answer 2

This is effectively a Riemann-Stieltjes integral.

Regardless, you should think of a differential of the form $\text{d}(f(x))$ as a variable substitution/parameterization, not a function.

In your example of $$\int_{z=0}^1\int_{\sin x=0}^y\cos x\ d\left(\sin x\right)dz$$

we make take $\sin(x)=u$ as a standard change of variables. We may also recall the property that for a function $g(x)$ that is continuously differentiable, $$\text{d}(g(x)) = g'(x)\text{ d}x$$ implying that $$\int f(x)\text{ d}(g(x)) = \int f(x)g'(x)\text{ d}x$$ This allows us to see that your example integral is simply just $$\int_{z=0}^1\int_{\text{placeholder}}^y\cos^2 x \text{ d}x\text{ d}z = \int_{\text{placeholder}}^y\cos^2 x \text{ d}x\cdot \int_{z=0}^1\text{ d}z = \int_{\text{placeholder}}^y\cos^2 x \text{ d}x$$

Now, for the placeholder bound, we don't need to calculate it, because if we work backwards from here with change of variables, we have $$u=\sin(x),\qquad \frac{\text{d}u}{\text{d}x} = \cos(x), \qquad x=\arcsin(u)$$ $$\implies \int_{u=0}^y\cos(\arcsin(u))\text{ d}u$$

I'm sure there's some domain shennanigans due to the inverse trig here, but for simplicity's sake, I'll ignore it.

This is what your integral effectively equals: https://www.wolframalpha.com/input?i=integrate+cos%28arcsinx%29+from+0+to+y