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

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

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integral from 0 to t of sin(3pit)

\int_{0}^{t} sin (3 π t) d t

\frac{1}{3 π} (- cos (3 π t) + 1)

Solution steps

\int_{0}^{t} sin (3 π t) d t

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

- cos (3 π t) + 1

= \frac{1}{3 π} (- cos (3 π t) + 1)

\int_{0}^{x} (3 t^{2} - t - 4) d t

\int_{- \infty}^{- 2} \frac{2}{x + 1} d x

\int_{0}^{s i n (t)} \frac{1}{9 - u ^{2}} d u

\int_{0}^{x} y d y

\int_{1}^{4} \frac{x - 3}{x} d x

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