What is the integral from 0 to pi+2i of cos(z/2) ?

The integral from 0 to pi+2i of cos(z/2) is 2sin((2i+pi)/2)

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integral from 0 to pi+2i of cos(z/2)

\int_{0}^{π + 2 i} cos (\frac{z}{2}) d z

2 sin (\frac{2 i + π}{2})

Solution steps

\int_{0}^{π + 2 i} cos (\frac{z}{2}) d z

Apply u-substitution

: \int_{0}^{\frac{π + 2 i}{2}} cos (u) \cdot 2 d u

= \int_{0}^{\frac{π + 2 i}{2}} cos (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}^{\frac{π + 2 i}{2}} cos (u) d u

Use the common integral:

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

= 2 [sin (u)]_{0}^{\frac{π + 2 i}{2}}

Compute the boundaries:

sin (\frac{2 i + π}{2})

= 2 sin (\frac{2 i + π}{2})

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What is the integral from 0 to pi+2i of cos(z/2) ?
The integral from 0 to pi+2i of cos(z/2) is 2sin((2i+pi)/2)