Why do we bother with $u$-substitution? This question has bothered me ever since I learned $u$-substitution (A note here: I have no formal education at this level, so I may definitely have missed something). The method is presented as an inverse concept to the chain rule and proved quite simply:
$$\frac{d}{dx}f(g(x)) = g'(x)f'(g(x)) \iff f(g(x)) = \int(g'(x)f'(g(x))\,dx$$
Let $u = g(x)$.  
Then $\dfrac{du}{dx}=g'(x) \implies du=g'(x)dx = u'dx$  
$$f(u) = \int u'f'(u) \, dx$$
$$f(u)=\int f'(u)\,du$$
Which is clearly true, leaving us with this new integration of $f'(x)dx$. But the system is somewhat unnatural, as in the end this entire process relies on the third line of the proof, where we manipulate infinitesimals. I've seen multiple questions on this site asking why it is okay to treat $\dfrac{dy}{dx}$ as a fraction and multiply out the denominator, and in the end it is unnecessary. Say we were to find $\int 2x\sin{x^2}dx$, intuitively we can do this as follows by directly finding $f(x)$ and $g(x)$:
$$f(x) = \text{outer function}=  \sin x$$
$$g(x) = \text{inner function with derivative} = x^2$$
Thus the original integral can be solved as $F(g(x))+C$ where F is an antiderivative of $f$, giving us our answer.
$$\int 2x\sin x^2\,dx = -\cos x^2 +C$$
But this is certainly not the way it is taught. I imagine there is an important reason for this, but what is that reason?
 A: I think the best justification is that it leads into the general context of substitutions for multiple integration. For example, it can be shown that the Jacobian of a polar transformation $$T(r,\theta)=(r\cos \theta,r\sin \theta)$$ is given by $$\begin{vmatrix}\cos\theta&\sin\theta\\ -r\sin\theta&r\cos\theta\end{vmatrix}=r(\cos^2\theta+\sin^2\theta)=r$$ This allows many integrals to be greatly simplified. For instance, the famous proof of the integral $$\int_0^\infty {e^{-x^2}\,\mathrm{d}x}$$ involves making the change of variables $$\iint{e^{-(x^2+y^2)}\,\mathrm{d}A}=\iint{re^{-r^2}\,\mathrm{d}r\,\mathrm{d}\theta}$$
U-substitutions are just the one-variable case of the much more general rule.
Also, even in the context of single-variable calculus some substitutions are unavoidable. For example, how do you propose to solve $$\int{\sqrt{1-x^2}\,\mathrm{d}x}$$ without making any substitutions? 
A: Any indefinite integral, which you can integrate to get a known function can be done just based on the fundamental theorem of calculus. All the other tricks like $u$-substitution, partial fractions, integration by parts only make this procedure easier, i.e., it helps us guessing the antiderivative.
A: Like everyone else has said, it makes life easier. Why it works: ∫f'(u)u' d*x* = ∫f'(u) d*u* (both of them equal to f(u), the first because of the chain rule, the second directly because of the fundamental theorem), so you can use u' d*x* = d*u* as a shorthand to rewrite ∫f'(u)u' d*x* in the formal method of u-substitution. 
