What is the integral from 0 to 1/2 of sin(2pix) ?

The integral from 0 to 1/2 of sin(2pix) is 1/pi

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integral from 0 to 1/2 of sin(2pix)

\int_{0}^{\frac{1}{2}} sin (2 π x) d x

\frac{1}{π}

Decimal

0.31830 \dots

Solution steps

\int_{0}^{\frac{1}{2}} sin (2 π x) d x

Apply u-substitution

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

= \int_{0}^{π} sin (u) \frac{1}{2 π} d u

Take the constant out:

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

= \frac{1}{2 π} \cdot \int_{0}^{π} sin (u) d u

Use the common integral:

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

= \frac{1}{2 π} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{2 π} \cdot 2

Simplify

= \frac{1}{π}

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What is the integral from 0 to 1/2 of sin(2pix) ?
The integral from 0 to 1/2 of sin(2pix) is 1/pi