Understanding substitution Hmm. I think I posted the wrong question before. Consider $I=\int_0^\pi \log(\sin(t)) \ dt$. Then $I\neq 0$ but the substitution $u=\sin(t)$ makes the end points equal so I end up having $I=0$. What am I doing wrong here? I don't quite understand the explanation at (Why should the substitution be injective when integrating by substitution?)
 A: Substitution: We start with $\int_a^b f(t)\, dt.$ If $g:[c,d]\to [a,b]$ is $C^1$ and $g(c) = a, g(d) =b ,$ then
$$\int_a^b f(t)\, dt = \int_c^d f(g(u))g'(u)\, du.$$
That's the theorem in its basic form. Now, when you tried "$u = \sin t$" above, you weren't following this recipe. That's where the problem is.
A: The basic question on substitution was already aptly addressed by zhw.  Another way of seeing this is that upon making the substitution $u=\sin x$ you are tacitly making the substitution $x = \arcsin u.$
But, the arcsine function has range $(-\pi/2, \pi/2)$, while $x$ spans $(0,\pi)$.  That is the reason for needing to split the integral into the sum of integrals over domains for which the appropriate inverse function can be identified.
The correct substitution is $x\to \arcsin x$ for $0<x<\pi/2$ and $x\to \pi -\arcsin x$ for $\pi/2<x<\pi$.  Then, 
$$\begin{align}
I&=\int_0^{\pi} \log (\sin x) dx\\
&=\int_0^{\pi/2} \log (\sin x) dx+\int_{\pi/2}^{\pi} \log (\sin x) dx\\
&=2\int_0^1 \frac{\log u}{\sqrt{1-u^2}}du
\end{align}$$
That said, we will proceed to evaluate the integral using the method of substitution.

Here, we will use a set of three simple substitutions that can be used in succession to evaluate this integral.  
First, we split the integral into two pieces as  
$$\begin{align}
I&=\int_0^{\pi} \log (\sin x) dx\tag 1\\
&=\int_0^{\pi/2} \log (\sin x) dx+\int_{\pi/2}^{\pi} \log (\sin x) dx\tag 2
\end{align}$$
Substitution 1:  We let $x\to \pi-x$ in the second integral of $(2)$ and find that
$$I=2 \int_0^{\pi/2} \log (\sin x) dx\tag 3$$
Substitution :  Next, we make the substitution $x \to 2x$ into $(1)$ and find that 
$$\begin{align}
I&=2 \int_0^{\pi/2}\log(\sin(2x)) dx\\
&=2\int_0^{\pi/2}\log 2 dx+2\int_0^{\pi/2}\log(\sin x) dx+2\int_0^{\pi/2}\log(\cos x) dx\\
&=\pi\log 2+I+2\int_0^{\pi/2}\cos x dx \tag 4
\end{align}$$
where we used $(2)$ to arrive at $(4)$.
Substitution 3:  Finally, we make the third substitution in the last integral of $(4)$.  There, letting $x \to \pi/2-x$ reveals that
$$\begin{align}
I&=\pi\log 2+I+2\int_0^{\pi/2}\log (\cos x) dx \\
&=\pi\log 2+I+2\int_0^{\pi/2}\log(\sin x) dx\\
&=\pi\log 2+2I
\end{align}$$
whereupon solving for $I$ yields $I=-\pi\log 2$.
