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Solving integrals/ sin(z)

Integral of sin(z) dz

Limits of integration:

from to
v

The graph:

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Piecewise:

The solution

You have entered [src] 
     p         
 I + -         
     2         
   /           
  |            
  |   sin(z) dz
  |            
 /             
 0
∫0p2+isin⁡(z) dz\int\limits_{0}^{\frac{p}{2} + i} \sin{\left(z \right)}\, dz 
Integral(sin(z), (z, 0, i + p/2))
Detail solution

  1. The integral of sine is negative cosine:

    ∫sin⁡(z) dz=−cos⁡(z)\int \sin{\left(z \right)}\, dz = - \cos{\left(z \right)}

  2. Add the constant of integration:

    −cos⁡(z)+constant- \cos{\left(z \right)}+ \mathrm{constant}

The answer is:

−cos⁡(z)+constant- \cos{\left(z \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                      
 |                       
 | sin(z) dz = C - cos(z)
 |                       
/
∫sin⁡(z) dz=C−cos⁡(z)\int \sin{\left(z \right)}\, dz = C - \cos{\left(z \right)}
Simplify
The answer [src] 
       /    p\
1 - cos|I + -|
       \    2/
1−cos⁡(p2+i)1 - \cos{\left(\frac{p}{2} + i \right)} 
=
= 
       /    p\
1 - cos|I + -|
       \    2/
1−cos⁡(p2+i)1 - \cos{\left(\frac{p}{2} + i \right)} 
1 - cos(i + p/2)

    Use the examples entering the upper and lower limits of integration.

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