What is the integral from 0 to 2pi of sin(mx) ?

The integral from 0 to 2pi of sin(mx) is (-cos(2pim)+1)/m

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integral from 0 to 2pi of sin(mx)

\int_{0}^{2 π} sin (m x) d x

\frac{- cos ( 2 πm ) + 1}{m}

Solution steps

\int_{0}^{2 π} sin (m x) d x

Apply u-substitution

: \int_{0}^{m \cdot 2 π} \frac{sin ( u )}{m} d u

= \int_{0}^{m \cdot 2 π} \frac{sin ( u )}{m} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{m} \cdot \int_{0}^{m \cdot 2 π} sin (u) d u

Use the common integral:

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

= \frac{1}{m} [- cos (u)]_{0}^{m \cdot 2 π}

Compute the boundaries:

- cos (2 πm) + 1

= \frac{1}{m} (- cos (2 πm) + 1)

Simplify

= \frac{- cos ( 2 πm ) + 1}{m}

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What is the integral from 0 to 2pi of sin(mx) ?
The integral from 0 to 2pi of sin(mx) is (-cos(2pim)+1)/m