$$\int \sin(2x+1)\,dx$$

Пусть $u=2x+1$, $du=2\,dx$: $$\int\sin(2x+1)\,dx=\frac12\int\sin u\,du=-\frac12\cos u+C=-\frac12\cos(2x+1)+C.$$

$$\boxed{-\dfrac12\cos(2x+1)+C}$$