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

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

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

\int_{0}^{\frac{1}{2}} sin (π t) d t

\frac{1}{π}

Decimal

0.31830 \dots

Solution steps

\int_{0}^{\frac{1}{2}} sin (π t) d t

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{π} [- cos (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{1}{π} \cdot 1

Simplify

= \frac{1}{π}

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