What is the integral from 0 to 3 of sin(pi/6 x) ?

The integral from 0 to 3 of sin(pi/6 x) is 6/pi

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integral from 0 to 3 of sin(pi/6 x)

\int_{0}^{3} sin (\frac{π}{6} x) d x

\frac{6}{π}

Solution steps

\int_{0}^{3} sin (\frac{π}{6} x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} sin (u) \frac{6}{π} d u

= \int_{0}^{\frac{π}{2}} sin (u) \frac{6}{π} d u

Take the constant out:

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

= \frac{6}{π} \cdot \int_{0}^{\frac{π}{2}} sin (u) d u

Use the common integral:

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

= \frac{6}{π} [- cos (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{6}{π} \cdot 1

Simplify

= \frac{6}{π}

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What is the integral from 0 to 3 of sin(pi/6 x) ?
The integral from 0 to 3 of sin(pi/6 x) is 6/pi