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

The integral from 0 to 2pi of 2a(sin(t/2)) is 8a

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

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

8 a

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{π} sin (u) \cdot 2 d u

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

Use the common integral:

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

= 2 a \cdot 2 [- cos (u)]_{0}^{π}

Simplify

= 4 a [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= 4 a \cdot 2

Simplify

= 8 a

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What is the integral from 0 to 2pi of 2a(sin(t/2)) ?
The integral from 0 to 2pi of 2a(sin(t/2)) is 8a