Changing limits of integration in trig

Question

I don't know why but I decided to solve this very easy integral by substituting $\sin x=t$ $$\int_{\pi/3}^{2\pi/3}xdx=\int_{x=\pi/3}^{x=2\pi/3}\frac{\arcsin t}{\sqrt{1-t^2}}dt$$ Since for $x=2\pi/3$ and $x=\pi/3$, $t=\sqrt{3}/2$, $$\int_{\sqrt{3}/2}^{\sqrt{3}/2}\frac{\arcsin t}{\sqrt{1-t^2}}dt$$ which, since each side of the limit is same, shoud be $0$, which obviously is not the answer.

What went wrong?

Answer 1

HINT

Note that the range of $\arcsin t$ is $[-\pi/2,\pi/2]$.

To fix you should set

$$\sin \left(x-\frac{\pi}2\right)=t \implies x=\arcsin t + \frac{\pi}2$$

Answer 2

The equation $\text{arcsin}(t)=\frac{2\pi}{3}$ has no solutions!

Arcsin is defined to be the inverse function of $\sin(t)$ on the interval $[-\pi/2,\pi/2]$. The equation $\text{arcsin}(t)=a$ has solution only if $-\pi/2\le a\le\pi/2$