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

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

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integral from 0 to 1 of sin(nx)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

- cos (n) + 1

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

Simplify

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

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