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

The integral from 0 to 2pi of sin(1/2)t is sin(1/2)*2pi^2

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integral from 0 to 2pi of sin(1/2)t

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

sin (\frac{1}{2}) \cdot 2 π^{2}

Decimal Notation

9.46348 \dots

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{2 π}

= sin (\frac{1}{2}) [\frac{t ^{2}}{2}]_{0}^{2 π}

Compute the boundaries:

2 π^{2}

= sin (\frac{1}{2}) \cdot 2 π^{2}

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