What is #int cosx-sin(x/2-pi) #?

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Answer 1 · 2016-11-27T18:12:50

#int (cos(x) - sin(x/2- pi))dx = sin(x) - 1/2cos(x/2) + C#

Given: #int (cos(x) - sin(x/2- pi))dx#

Use the identity, #sin(A - B) = sin(A)cos(B) - cos(A)sin(B)#:

#sin(x/2 - pi) = sin(x/2)cos(pi) - cos(x/2)sin(pi) #

#sin(x/2 - pi) = -sin(x/2) #

Substitute into the integrand:

#int (cos(x) - (-sin(x/2)))dx#

#int (cos(x) + sin(x/2))dx = sin(x) - 1/2cos(x/2) + C#