What is the integral from 0 to 1/z of sin(z) ?

The integral from 0 to 1/z of sin(z) is -cos(1/z)+1

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integral from 0 to 1/z of sin(z)

\int_{0}^{\frac{1}{z}} sin (z) d z

- cos (\frac{1}{z}) + 1

Solution steps

\int_{0}^{\frac{1}{z}} sin (z) d z

Use the common integral:

\int sin (z) d z = - cos (z)

= [- cos (z)]_{0}^{\frac{1}{z}}

Compute the boundaries:

- cos (\frac{1}{z}) + 1

= - cos (\frac{1}{z}) + 1

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What is the integral from 0 to 1/z of sin(z) ?
The integral from 0 to 1/z of sin(z) is -cos(1/z)+1