How do you find the definite integral for: #cos(x)# for the intervals #[-pi, pi]#?

Answer 1

#int_(-pi)^pi cosxdx=[sinx]_(-pi)^(pi)=sinpi-sin(-pi)=0#.

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Answer 2

To find the definite integral of cos(x) over the interval [-π, π], you can use the properties of trigonometric functions and integration techniques. The integral of cos(x) from -π to π is equal to the integral of cos(x) from 0 to π plus the integral of cos(x) from -π to 0.

Using the property of evenness of cosine function, the integral of cos(x) from -π to 0 is equal to the integral of cos(x) from 0 to π. Therefore, the definite integral of cos(x) over the interval [-π, π] is twice the integral of cos(x) from 0 to π.

Integrating cos(x) from 0 to π yields sin(x), evaluated from 0 to π. So, sin(π) - sin(0).

Since sin(π) is 0 and sin(0) is 0, the definite integral of cos(x) over the interval [-π, π] is 0.

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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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