What is the integral from 0 to 2pi of sin(t/2) ?

The integral from 0 to 2pi of sin(t/2) is 4

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integral from 0 to 2pi of sin(t/2)

\int_{0}^{2 π} sin (\frac{t}{2}) d t

4

Solution steps

\int_{0}^{2 π} sin (\frac{t}{2}) d t

Apply u-substitution

: \int_{0}^{π} sin (u) \cdot 2 d u

= \int_{0}^{π} sin (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int_{0}^{π} sin (u) d u

Use the common integral:

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

= 2 [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= 2 \cdot 2

Simplify

= 4

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What is the integral from 0 to 2pi of sin(t/2) ?
The integral from 0 to 2pi of sin(t/2) is 4