U-Substitution Intuition
Two ways to look at it:
First, the indefinite integral: $\int F'(x)\,dx = F(x)+C$ is the antiderivative. In this viewpoint, the substitution rule is just the chain rule written backwards: $\int F'(g(x))\cdot g'(x)\,dx = F(g(x))+C$.
Second, the definite integral as the area problem; $\int_a^b f(x)\,dx$ is the area under the graph of $f$ between $a$ and $b$. Here, a substitution will transform the interval we integrate over, and we'll need to stretch the function vertically in order to keep the area the same:
Our transformation stretches a small horizontal segment $dx$ to $du$, multiplying by $\frac{du}{dx}$. In order to keep the same area, we have to multiply the function values by the reciprocal $\frac{dx}{du}$. The example here $(f(x)=\sqrt{1-x^2},u=\sin x)$ has $\frac{du}{dx} > 1$, so the transformed graph is shorter and wider in this case.
If we slice up the whole interval this way, each slice of the area under the transformed graph has the same area as the corresponding slice of the area under the original graph. Sum them up, and the areas are the same.
Believe it or not, u-substantiation is nothing more than doing the chain rule backwards. In other words, when you're doing u-substitution, you're going in the opposite direction. Take a look at the following example where we're about to take the derivative of the function $u(x)$:
$$\int f(u)\frac{du}{dx}\,dx=h(x)\Longleftrightarrow h'(x)=f(u)\frac{du}{dx}\\$$
I hope you agree that what we have on the left is equivalent to what we have on the right. Now, let's go further back in the opposite direction. We have differentiated $h(x)$ to get $f(u)$ and $\frac{du}{dx}$ is just about to appear, but let's freeze things for a second here:
$$h'(x)=f(u)\implies\int f(u)\,du=h(x)$$
Of course, $u$ is a function of $x$ there. What we've got on the left side is a new integral that should be integrated with respect to the variable $u$. The idea here is that the new integral we got will hopefully be easier to integrate than the one we had at the beginning. So, in short, when you're doing u-substitution, you're transforming the original problem into an incomplete chain rule problem.
Let's do a simple but concrete example to illustrate this process:
$$ \int(x+1)^2\,dx=h(x)\Longleftrightarrow h'(x)=(x+1)^2\\ \int(x+1)^2\cdot1\,dx=h(x)\Longleftrightarrow h'(x)=(x+1)^2\cdot1\\ \int(x+1)^2\frac{d}{dx}(x+1)\,dx=h(x)\Longleftrightarrow h'(x)=(x^2+1)^2\frac{d}{dx}(x+1) $$
Let's replace $x+1$ with $u$ to make things easier to read:
$$\int u^2\frac{du}{dx}\,dx=h(x)\Longleftrightarrow h'(x)=u^2\frac{du}{dx}$$
Let's take one more step back. We have differentiated $h(x)$ to get $u^2$ and it's just about time to take the derivative of $u(x)$, but we freeze things instead:
$$\int u^2\,du=h(x)\Longleftrightarrow h'(x)=u^2$$
And now, the integral that we've got is easy to differentiate because it's one of those elementary table integrals that we all know how to do:
$$ \int u^2\,du=\frac{u^3}{3}+C=\frac{(x+1)^3}{3}+C. $$