What is the integral from 0 to t of sin(3x) ?

The integral from 0 to t of sin(3x) is 1/3 (-cos(3t)+1)

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integral from 0 to t of sin(3x)

\int_{0}^{t} sin (3 x) d x

\frac{1}{3} (- cos (3 t) + 1)

Solution steps

\int_{0}^{t} sin (3 x) d x

Apply u-substitution

: \int_{0}^{3 t} sin (u) \frac{1}{3} d u

= \int_{0}^{3 t} sin (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{3 t} sin (u) d u

Use the common integral:

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

= \frac{1}{3} [- cos (u)]_{0}^{3 t}

Compute the boundaries:

- cos (3 t) + 1

= \frac{1}{3} (- cos (3 t) + 1)

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What is the integral from 0 to t of sin(3x) ?
The integral from 0 to t of sin(3x) is 1/3 (-cos(3t)+1)