What is the integral from 0 to t of 1/2 sin(2x) ?

The integral from 0 to t of 1/2 sin(2x) is 1/4 (-cos(2t)+1)

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integral from 0 to t of 1/2 sin(2x)

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

\frac{1}{4} (- cos (2 t) + 1)

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{2 t} : \frac{1}{4} [- cos (u)]_{0}^{2 t}

= \frac{1}{4} [- cos (u)]_{0}^{2 t}

Compute the boundaries:

- cos (2 t) + 1

= \frac{1}{4} (- cos (2 t) + 1)

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What is the integral from 0 to t of 1/2 sin(2x) ?
The integral from 0 to t of 1/2 sin(2x) is 1/4 (-cos(2t)+1)