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

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

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

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

- 2 cos (\frac{2 i + π}{2})

Solution steps

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

Apply u-substitution

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

= \int_{\frac{π}{2}}^{\frac{π + 2 i}{2}} 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_{\frac{π}{2}}^{\frac{π + 2 i}{2}} sin (u) d u

Use the common integral:

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

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

Compute the boundaries:

- cos (\frac{2 i + π}{2})

= 2 (- cos (\frac{2 i + π}{2}))

Simplify

= - 2 cos (\frac{2 i + π}{2})

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