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

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

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integral from 0 to 1 of sin(wx)

\int_{0}^{1} sin (w x) d x

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

Solution steps

\int_{0}^{1} sin (w x) d x

Apply u-substitution

: \int_{0}^{w} \frac{sin ( u )}{w} d u

= \int_{0}^{w} \frac{sin ( u )}{w} d u

Take the constant out:

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

= \frac{1}{w} \cdot \int_{0}^{w} sin (u) d u

Use the common integral:

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

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

Compute the boundaries:

- cos (w) + 1

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

Simplify

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

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