What is the integral from 0 to t of sin(2u) ?

The integral from 0 to t of sin(2u) is 1/2 (-cos(2t)+1)

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integral from 0 to t of sin(2u)

\int_{0}^{t} sin (2 u) d u

\frac{1}{2} (- cos (2 t) + 1)

Solution steps

\int_{0}^{t} sin (2 u) d u

Apply u-substitution

: \int_{0}^{2 t} sin (v) \frac{1}{2} d v

= \int_{0}^{2 t} sin (v) \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2 t} sin (v) d v

Use the common integral:

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

= \frac{1}{2} [- cos (v)]_{0}^{2 t}

Compute the boundaries:

- cos (2 t) + 1

= \frac{1}{2} (- cos (2 t) + 1)

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What is the integral from 0 to t of sin(2u) ?
The integral from 0 to t of sin(2u) is 1/2 (-cos(2t)+1)