What is the integral from 2pi to infinity of sin(θ) ?

The integral from 2pi to infinity of sin(θ) is diverges

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integral from 2pi to infinity of sin(θ)

\int_{2 π}^{\infty} sin (θ) d θ

diverges

Solution steps

\int_{2 π}^{\infty} sin (θ) d θ

Compute the indefinite integral:

\int sin (θ) d θ = - cos (θ) + C

Compute the boundaries:

\int_{2 π}^{\infty} sin (θ) d θ =

diverges

= diverges

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What is the integral from 2pi to infinity of sin(θ) ?
The integral from 2pi to infinity of sin(θ) is diverges