Solved on Jan 23, 2024

Compute the indefinite integrals with constant of integration $C$ : $\int -\sin(t) dt$ , $\int -\cos(t) dt$ .

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 $t$ .
3. We will use $C$ to represent the constant of integration.

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

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Compute the indefinite integrals with constant of integration $C$ : $\int -\sin(t) dt$ , $\int -\cos(t) dt$ .

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 $t$ .
3. We will use $C$ to represent the constant of integration.

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

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

STEP 1

Assumptions1. 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 t.3. We will use C 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∫sin(t)dt=−cos(t)+C ∫cos⁡(t)dt=sin⁡(t)+C\int \cos(t) dt = \sin(t) + C∫cos(t)dt=sin(t)+C

STEP 3

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

STEP 4

Use the basic antiderivative formula for sin⁡(t) \sin(t) sin(t), but remember to include the negative sign:∫−sin⁡(t)dt=−(−cos⁡(t)+C)\int -\sin(t) dt = -(-\cos(t) + C)∫−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∫−sin(t)dt=cos(t)−CHowever, since C C C is an arbitrary constant, we can absorb the negative sign into it, as −C -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∫−sin(t)dt=cos(t)+C

STEP 7

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

STEP 8

Use the basic antiderivative formula for cos⁡(t) \cos(t) cos(t), but remember to include the negative sign:∫−cos⁡(t)dt=−(sin⁡(t)+C)\int -\cos(t) dt = -(\sin(t) + C)∫−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∫−cos(t)dt=−sin(t)−CAgain, since C C C is an arbitrary constant, we can absorb the negative sign into it.

STEP 10

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Compute the indefinite integrals with constant of integration $C$ : $\int -\sin(t) dt$ , $\int -\cos(t) dt$ .

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 $t$ .
3. We will use $C$ to represent the constant of integration.

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

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

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

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

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

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

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

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