Evaluate the integral

\(\int_{-2}^{\infty}\sin x dx = [\int \sin xdx]_{-2}^{\infty}\)

\(=[-\cos(x)]_{-2}^{\infty}=[-\cos(\infty)+\cos(-2)]=[-\cos(\infty)+\cos 2]\)

Now, note that

\(\lim_{x\rightarrow \infty}\cos(x)\) doesn't exist, i.e. − \(\cos(\infty)\) doesn't exist.

\(\Rightarrow [-\cos(\infty)+\cos 2]\) doesn't exist

Thus, the integral

\(\int_{-2}^{\infty}\sin(x)dx\) is divergent

\(\int_{-2}^{\infty}\sin x dx = [\int \sin xdx]_{-2}^{\infty}\)

\(=[-\cos(x)]_{-2}^{\infty}=[-\cos(\infty)+\cos(-2)]=[-\cos(\infty)+\cos 2]\)

Now, note that

\(\lim_{x\rightarrow \infty}\cos(x)\) doesn't exist, i.e. − \(\cos(\infty)\) doesn't exist.

\(\Rightarrow [-\cos(\infty)+\cos 2]\) doesn't exist

Thus, the integral

\(\int_{-2}^{\infty}\sin(x)dx\) is divergent