What 's wrong with this integral substitution? What's wrong with the following solution?
Let $t=\sin x,$ then $x=\arcsin t.$ By substitution method we have the following:
$\int_0^{2\pi}\sin^2x \rm{dx}=\int_0^0 \frac{t^2}{\sqrt{1-t^2}}\rm{dt}=0$.
However, we know that  $\int_0^{2\pi}\sin^2x \rm{dx}=\int_0^{2\pi}\frac{1-\cos 2x}{2} \rm{dx}=\pi.$
I can't find the problem.
Thanks for any suggestions.
 A: The substitution $$x=\arcsin t$$ is valid for $$-\pi/2\le x\le \pi/2$$ where $ \arcsin t$ is defined. 
A: I will give you another example (much more extreme) so you can see plainly why is failing here.
If you consider $\int_{-1}^1 (1-t^2) dt$ and you try to perform the substitution $1-t^2=x$, then $-2t dt = dx$, so $dt = \frac{-1}{2\sqrt{1-x}} dx$.
$$\int_{-1}^1 (1-t^2) dt = \int_{0}^0 x \frac{-dx}{2\sqrt{1-x}} = 0$$
Which makes no sense since the integral on the left is clearly non zero. The problem is that the substitution we made is not bijective on these intervals. The same happens with yours.
A: The reason why the lower and upper limits turn into the same value $0$ after changing variable is due to the lack of bijectivity of the substitution, even though the calculation is incorrect. Bijectivity in changing the variable is not a strong requirement, but it needs to be careful in calculation when a non-bijective substitution is applied.
Correct calculation using the same substitution is as follows: From $1-t^2 = x$, $t^2 = 1-x \ge 0$ and so
$$t = \begin{cases}
\sqrt{1-x} & \text{for $t \ge 0$,} \\
-\sqrt{1-x} & \text{for $t < 0$.}
\end{cases}$$
with
$$dt = \begin{cases}
-\frac{1}{2\sqrt{1-x}}\,dx & \text{for $t > 0$} \\
\frac{1}{2\sqrt{1-x}}\,dx & \text{for $t < 0$.}
\end{cases}$$
Hence
\begin{align*}
\int_{-1}^{1}(1-t^2)\,dt 
  & = \int_{-1}^{0} (1-t^2)\,dt + \int_0^1 (1-t^2)\,dt \\
  & = \int_0^1 x\left(\frac{1}{2\sqrt{1-x}}\right)\,dx 
      + \int_1^0 x\left(\color{red}{-}\frac{1}{2\sqrt{1-x}}\right)\,dx\\
  & = \int_0^1 \frac{x}{2\sqrt{1-x}}\,dx 
      + \int_\color{red}{0}^\color{red}{1} \frac{x}{2\sqrt{1-x}}\,dx \\
  & = \int_0^1 \frac{x}{\sqrt{1-x}}\,dx 
    = \lim_{\varepsilon \rightarrow 1^-} \int_0^{\varepsilon} \frac{x}{\sqrt{1-x}}\,dx\\
  & = \lim_{\varepsilon \rightarrow 1^-} \left[\left.-\frac{2}{3}\sqrt{1-x}(x+2)\right|_0^\varepsilon\right] \\
  & = \lim_{\varepsilon \rightarrow 1^-} \left[\left(-\frac{2}{3}\sqrt{1-\varepsilon}(\varepsilon+2)\right) - \left(-\frac{4}{3}\right)\right] \\ 
  & = 0 - \left(-\frac{4}{3}\right) = \frac{4}{3}.
\end{align*}
