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

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

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integral from 0 to x/3 of sin(3t)

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

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

Solution steps

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

Apply u-substitution

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

Use the common integral:

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

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

Compute the boundaries:

- cos (x) + 1

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

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