Why $\int \limits _0 ^1 \log x \Bbb d x \ne \int \limits _0 ^{2 \pi} \log \sin x \Bbb d x$? From what I know integration is the summation of all infinitesimally small identical objects.
So won't the integration in both the cases be the sum of $\log(0), \dots, \log(1)$ because we're pretty much providing the same input to the logarithmic function?
According to Wolfram Alpha the results are different.
 A: I will suppose that you want to understand why $\int_{0}^{\pi/2}\log\sin x\,dx \neq \int_{0}^{1}\log(x)\,dx$.
You are probably thinking that the integral should be the same because the values through which the function $f(x)=\log\sin x$ goes on the interval $[0, \pi/2]$ are the same as the values through which the function $g(x)=\log x$ goes on the interval $[0,1]$ and both functions are attaining the same values exactly once.
But if you look at the integral as the area of the curve (although in this two examples the area will be negative because graphs of the functions are below $x$-axis, but it does not matter for the discussion) then you have here two different curves, the curve $x\to (x,\log\sin x)$ on the interval $[0,\pi/2]$ and the curve $x\to (x,\log x)$ on the interval $[0,1]$ and although the $y$-coordinates of both curves will represent the same set of values the curves does not attain the same $y$-coordinate values at the equal $x$-coordinates (except $x=0$) so we are talking about two different curves so the integrals will also be different.
A: Why should they be equal? You're integrating two very different curves: 
These are your functions, clearly you're integrating something completely different. 

But beyond this (because there are different functions whose integrals give the same results) the $\sin(x)$ functions behaves really different from the $x$ function. 
When you integrate by substitutions, you have to take into account the so called Jacobian of the transformation which changes everything.
If you want, you can always try to solve this problem: "find those values $a$ and $b$ for which the integral of $\log(\sin(x))$ gives you $-1$".
Hint
$$\int_a^b \log\sin(x)\ \text{d} x = -x\log\left(1 - e^{2ix}\right) + x\log\left(\sin(x)\right) + \frac{1}{2}i\left(x^2 + \text{PolyLog}\left[2, e^{2ix}\right]\right)\bigg|_{a}^b$$
A: If you are attempting a substitution, I think the substitution is wrong. Let $y=\sin(x)$. The integral limits become $0$ and $\sin(1)$. Then we get
\begin{equation*}
\int ^1_0 \ln(x)dx= \int^{\sin(1)}_{0} \frac{\ln(\sin(y))}{\cos(y)}dy.
\end{equation*}
If you integrate $\ln(x)$ from $0$ to $1$, we can use integration be parts to get $-1$ by treating $\ln(x)$ as $1\times \ln(x)$. 
