TITLE: Substitution Makes the Integral Bounds Equal
QUESTION [15 upvotes]: This seems like a really basic calculus question, which is a tad embarrassing since I'm a graduate student, but what does it mean when a substitution in a definite integral makes the bounds the same? For example, if we have some function of $\sin(x)$:
$$\int_0^{\pi} f(\sin(x)) \,\mathrm{d}x$$
If we make the substitution $u = \sin(x)$, then $du = \cos(x)\,\mathrm{d}x$, we find
$$\int_{\sin(0)}^{\sin(\pi)} \frac{f(u)}{\cos(x)} \,\mathrm{d}u 
= \int_0^0 \frac{f(u)}{\sqrt{1-u^2}} \,\mathrm{d}u$$
This would imply that the integral is zero. Is this always the case? For another example (more relevant to the problem I'm actually trying to solve) consider
$$\int_{-b}^{b} \frac{1}{\sqrt{x^2 + a^2}}\,\mathrm{d}x$$
Clearly this can be solved using a trigonometric substitution to get $2\operatorname{arcsinh}(b)$, but what if I substituted $u = \sqrt{x^2 + a^2}$? Then
$$\mathrm{d}u = \frac{x\,\mathrm{d}x}{\sqrt{x^2 + a^2}} \implies \mathrm{d}x = \frac{u\,\mathrm{d}u}{x} = \frac{u\, \mathrm{d}u}{\sqrt{u^2 - a^2}},$$
so the integral becomes
$$\int_{\sqrt{b^2 + a^2}}^{\sqrt{b^2 + a^2}} 
\frac{1}{\sqrt{u^2 - a^2}}\,\mathrm{d}u$$
This integral seems to be zero, which is not the case for the integral before the substitution. What's going on here? Does this just mean that these substitutions are not valid?

REPLY [3 votes]: The first integral is NOT zero! Let $f$ be the identity function, for example.
In the second integral it is wrong to say that $x=\sqrt{u^2-a^2}$ for all values of $x$. In the first integral, the same: $cos(x)=\sqrt{1-u^2}$ is not true for all values of $x$.
When the substitution is not injective, problems arise when you try to express the integrand in terms of the new variable, as it can be seen from this examples. So always split the integration domain so that there is injectivity in each part.