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

Answer 1

#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) #

Put in place of the integrand:

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

The expression ( \text{int} \cos(x) - \sin\left(\frac{x}{2} - \pi\right) ) is not clear. If "int" is meant to represent the integer part of a number, then it would depend on the specific value of ( x ). If it represents an integral, then it would need limits of integration. Please clarify the expression or provide additional context.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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