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?
For the second integral, note that the substitution $u=\sqrt{x^2+a^2}$ implies: $$ u\ge 0 \quad \mbox{and}\quad x=\pm\sqrt{u^2-a^2} $$ so:
$$ dx=\frac{udu}{\sqrt{u^2-a^2}} \mbox{for}\quad x \ge 0 $$
$$ dx=\frac{udu}{-\sqrt{u^2-a^2}} \mbox{for}\quad x < 0 $$
and the integral splits in two parts as $\int_{-b}^0 +\int_0^b$. This gives the correct result.
We have an analogous situation for the first integral with the substitution $$ u=\sin x \qquad \cos x=\pm \sqrt{1-u^2} $$