The easiest way is to write your function as $(\sin t)(1-\cos^2 t)=\sin t -\sin t\; \cos^2 t$. For the second part, make the substitution $u=\cos t$.
A similar idea works for anything of the shape $(\sin^m t)(\cos^n t)$ where $m$ and $n$ are integers at least one of which is odd.
For example, to integrate $\sin^2 t \;\cos^5 t$, rewrite as $\sin^2 t \;\cos^4 t\; \cos t$. Then express $\cos^4 t$ as $(1-\sin^2 t)^2$, and make the substitution $u=\sin t$.
Comment: Your integration by parts would also have solved the problem. Let $u=\sin^2 t$ and $dv=\sin t\,dt$. Then $du=2\sin t\,\cos t$, and we can take $v=-\cos t$.
So we have
$$\int \sin^3 t\,dt =-\cos t \, \sin^2 t + \int 2\sin t\, \cos^2 t\, dt.$$
The remaining integral yields to the substitution $w=\cos t$.
But let's pretend we didn't notice this. In the integral, replace $\cos^2 t$ by $1-\sin^2 t$. After a little while, we arrive at
$$\int \sin^3 t\,dt=-\cos t\, \sin^2 t +2\cos t -\int 2\sin^3 t\,dt.$$
Now it really looks as if we are stuck in the loop you referred to. The above equation seems to say "I will tell you the answer if you give me the answer." However, let $I=\int \sin^3 t\,dt$. Then we have
$$I=-\cos t\,\sin^2 t +2\cos t -2I.$$
It follows that $3I=-\cos t\,\sin^2 t +2\cos t$. Now we can divide by $3$ to get $I$. Don't forget to add the arbitrary constant of integration.
The above trick comes up surprisingly often. When it looks as if we are going in circles, we usually are going in circles. But not always!