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Question Compute the indefinite integrals with constant of integration CC: ∫−sin⁡(t)dt\int -\sin(t) dt, ∫−cos⁡(t)dt\int -\cos(t) dt.

Studdy Solution

STEP 1

Assumptions
1. We are working with indefinite integrals, which means we are finding the antiderivative of the given functions.
2. The variable of integration is t t .
3. We will use C C to represent the constant of integration.

STEP 2

Recall the basic antiderivative formulas for trigonometric functions:
∫sin⁡(t)dt=−cos⁡(t)+C\int \sin(t) dt = -\cos(t) + C ∫cos⁡(t)dt=sin⁡(t)+C\int \cos(t) dt = \sin(t) + C

STEP 3

Compute the indefinite integral of the first function:
∫−sin⁡(t)dt\int -\sin(t) dt

STEP 4

Use the basic antiderivative formula for sin⁡(t) \sin(t) , but remember to include the negative sign:
∫−sin⁡(t)dt=−(−cos⁡(t)+C)\int -\sin(t) dt = -(-\cos(t) + C)

STEP 5

Simplify the expression by distributing the negative sign:
∫−sin⁡(t)dt=cos⁡(t)−C\int -\sin(t) dt = \cos(t) - C
However, since C C is an arbitrary constant, we can absorb the negative sign into it, as −C -C is also an arbitrary constant.

STEP 6

Write the final answer for the first integral:
∫−sin⁡(t)dt=cos⁡(t)+C\int -\sin(t) dt = \cos(t) + C

STEP 7

Compute the indefinite integral of the second function:
∫−cos⁡(t)dt\int -\cos(t) dt

STEP 8

Use the basic antiderivative formula for cos⁡(t) \cos(t) , but remember to include the negative sign:
∫−cos⁡(t)dt=−(sin⁡(t)+C)\int -\cos(t) dt = -(\sin(t) + C)

STEP 9

Simplify the expression by distributing the negative sign:
∫−cos⁡(t)dt=−sin⁡(t)−C\int -\cos(t) dt = -\sin(t) - C
Again, since C C is an arbitrary constant, we can absorb the negative sign into it.

STEP 10

Write the final answer for the second integral:
∫−cos⁡(t)dt=−sin⁡(t)+C\int -\cos(t) dt = -\sin(t) + C
The solutions to the indefinite integrals are:
∫−sin⁡(t)dt=cos⁡(t)+C∫−cos⁡(t)dt=−sin⁡(t)+C \begin{array}{l} \int-\sin (t) d t = \cos(t) + C \\ \int-\cos (t) d t = -\sin(t) + C \end{array}

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