p I + - 2 / | | sin(z) dz | / 0
Integral(sin(z), (z, 0, i + p/2))
The integral of sine is negative cosine:
Add the constant of integration:
The answer is:
/ | | sin(z) dz = C - cos(z) | /
/ p\ 1 - cos|I + -| \ 2/
=
/ p\ 1 - cos|I + -| \ 2/
1 - cos(i + p/2)
Use the examples entering the upper and lower limits of integration.