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

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

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integral from 0 to t of sin(2s)

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

- cos (2 t) + 1

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

\int \frac{2}{10 + 6 x + x ^{2}} d x

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\int \frac{sin ^{2} ( x ) cos ( 2 x )}{2} d x

\int_{1}^{2} \frac{( 2 + 2 x - x ^{2} )}{8} d x

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