# why the ending symbol dx is also changed into a function of d theta and then multiplied by the integral after substitution? [duplicate]

This is a question about trig substitution used in integrals.

Because it is difficult to solve an integral when there is radical in it, we use a trig function of theta to substitute x from the original integral.

The only thing I don't understand is why the ending symbol dx, which indicates that we are integrating the function with respect to the "x" variable, is also changed into a function of d theta and then multiplied by the integral after substitution.

I think after substitution, we can just rewrite dx into d theta.

• if $\theta=f(x)$, then $d\theta=f'(x)dx$ Mar 16 at 12:28

$$dx$$ and $$d\theta$$ are not the same thing. For example, if $$x = \sin \theta$$, then $$\frac{dx}{d\theta} = \cos \theta$$ and so $$dx = \cos {\theta} d\theta$$.
• @mirthspritzultyrobscurantism So there are two ways of resolving this. One is simply saying that, by the fundamental theorem of calculus, and the chain rule, we have $\int_a^b g'(x)f(g(x))\textrm{d}x=\int_{g(a)}^{g(b)} f(u)\textrm{d}u$. This is a borderline trivial and completely rigorous statement. One short-hand for this could be to write "$u=g(x)$ implies $\textrm{d}u=\frac{du}{dx}\textrm{d}x$ since that's sort of how the above look if you think all you're doing is integrating $f$. And this heuristic will always yield correct substitutions. Mar 16 at 20:17
• Another is appealing to measure theory, stating that $\textrm{d}x$ is a measure (the ordinary Lesbegue measure) and $\textrm{d}u$ is the measure such that $\textrm{d}u([a,b])=u(b)-u(a)$. In this formulation, $\textrm{d}u=\frac{du}{dx} \textrm{d}x$ is true as a statement about measures with densities and can be proven rigorously. One should note, though, that $\frac{du}{dx}$ is a function and not a measure, so although the notation is suggestive, the result isn't literally due to "fractions cancelling out". Mar 16 at 20:21