Solve the equations and show all the solutions in the first quadrant 3\ta

Let P(x, y) be the terminal point on the unit circle determined by t. Then $\sin t=,\cos t=,\text{and}\tan t=$.

Why ${\displaystyle {\int }_0^{2\pi }{\sin }^{2}xdx\ne {\int }_0^{2\pi }\sin x\sin nxdx\to n=1}$
By orthogonality, we know that
${\int }_0^{2\pi }\sin mx\sin nxdx=\pi$ if $m=n$ and 0 otherwise. Nevertheless, when I am calculating it as follows, I get 0.
${\displaystyle {\int }_0^{2\pi }\sin x\sin nxdx=\frac{1}{2}{[\frac{\sin \left[(1-n)x\right]}{1-n}-\frac{\sin \left[(1+n)x\right]}{1+n}]}_0^{2\pi }}$
But the above quantity is 0 even if n=1. Shouldn't it be ${\displaystyle \pi }$?