What is the integral from 0 to x of sin(8u) ?

The integral from 0 to x of sin(8u) is 1/8 (-cos(8x)+1)

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integral from 0 to x of sin(8u)

\int_{0}^{x} sin (8 u) d u

\frac{1}{8} (- cos (8 x) + 1)

Solution steps

\int_{0}^{x} sin (8 u) d u

Apply u-substitution

: \int_{0}^{8 x} sin (v) \frac{1}{8} d v

= \int_{0}^{8 x} sin (v) \frac{1}{8} d v

Take the constant out:

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

= \frac{1}{8} \cdot \int_{0}^{8 x} sin (v) d v

Use the common integral:

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

= \frac{1}{8} [- cos (v)]_{0}^{8 x}

Compute the boundaries:

- cos (8 x) + 1

= \frac{1}{8} (- cos (8 x) + 1)

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What is the integral from 0 to x of sin(8u) ?
The integral from 0 to x of sin(8u) is 1/8 (-cos(8x)+1)