I'm going through the MIT open calculus course, and in one of the lectures (19-28min marks) the professor uses the trig substitution $x = \tan \theta$ to find the integral of $\frac{dx}{x^2 \sqrt{1+x^2}}$.
His answer: $-\csc(\arctan x) + c$, which he shows is equivalent to $-\frac{1+x^2}{x} + c$ by drawing a right triangle on the blackboard.
I get the math behind each step of it, but I can't wrap my head around why that equivalence works. We just used an arbitrary $x = \tan \theta$ substitution, where $\theta$ moves differently than x does, and the expression $-\frac{1+x^2}{x} + c$ by itself doesn't know anything about trigonometry. But I type both into Excel for a bunch of different x values, and obviously they are equivalent.
I guess I'm not really sure what my question is here, but I could just use some perspective. It just seems like substituting ANY function in for x then integrating it shouldn't work, especially when crossing into polar coordinates.
Best Answer
This has to do with the fact that although $1/\sqrt{1+x^2}$ is difficult to integrate, $$1/\sqrt{1+\tan^2 t}= 1/\sqrt{1/\cos^2 t}=\cos t$$ And together with $dx = dt/\cos^2 t $, leads to a very simple integral. In essence, any substitution "works", in the sense that all lead to the same numerical value, but only some of these lead to an easier integration.
As to the general idea behind substitution, take the easy example of $t = ax$: $$\int_0^1 f(x) dx = \int_0^a \frac{f(x/a)}{a} dt $$ What's happening here? If you think geometrically, we just stretched the shape along the $x$-axis, so it's width is now $a$ instead of $1$. If we were to just naively substitute $t=ax$, the area would be $a$ times larger! and that is why we need to correct for this "stretching" by dividing by $dx/dt=a$.
As you mentioned, in the more general case the stretching is no longer uniform, but you can look at the new integral as a sum of differently stretched pieces, each with it's own correction factor $$dt/dx \neq \rm{const}$$.
By way of example, take a look at the integral over the half circle $\sqrt{1-x^2}$, with the transformation $x=\sin(t)$. You can see how the uniformly distributed widths in $x$ get "squeezed" differently in $t$: