Question

Evaluate the ff, improper integrals. \int_{-2}^{\infty}\sin x dx

Applications of integrals
ANSWERED
asked 2021-05-23
Evaluate the ff, improper integrals.
\(\int_{-2}^{\infty}\sin x dx\)

Expert Answers (1)

2021-05-24
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
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