Calculus Examples

$19.21 \sin (1.7 t + 0.3) - 16.32 \cos (1.7 t + 0.3)$

Step 1

Write $19.21 \sin (1.7 t + 0.3) - 16.32 \cos (1.7 t + 0.3)$ as a function.

$f (t) = 19.21 \sin (1.7 t + 0.3) - 16.32 \cos (1.7 t + 0.3)$

Step 2

The function $F (t)$ can be found by finding the indefinite integral of the derivative $f (t)$ .

$F (t) = \int f (t) d t$

Step 3

Set up the integral to solve.

$F (t) = \int 19.21 \sin (1.7 t + 0.3) - 16.32 \cos (1.7 t + 0.3) d t$

Step 4

Split the single integral into multiple integrals.

$\int 19.21 \sin (1.7 t + 0.3) d t + \int - 16.32 \cos (1.7 t + 0.3) d t$

Step 5

Since $19.21$ is constant with respect to $t$ , move $19.21$ out of the integral.

$19.21 \int \sin (1.7 t + 0.3) d t + \int - 16.32 \cos (1.7 t + 0.3) d t$

Step 6

Let $u_{1} = 1.7 t + 0.3$ . Then $d u_{1} = 1.7 d t$ , so $\frac{1}{1.7} d u_{1} = d t$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 6.1

Let $u_{1} = 1.7 t + 0.3$ . Find $\frac{d u_{1}}{d t}$ .

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Step 6.1.1

Differentiate $1.7 t + 0.3$ .

$\frac{d}{d t} [1.7 t + 0.3]$

Step 6.1.2

By the Sum Rule, the derivative of $1.7 t + 0.3$ with respect to $t$ is $\frac{d}{d t} [1.7 t] + \frac{d}{d t} [0.3]$ .

$\frac{d}{d t} [1.7 t] + \frac{d}{d t} [0.3]$

Step 6.1.3

Evaluate $\frac{d}{d t} [1.7 t]$ .

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Step 6.1.3.1

Since $1.7$ is constant with respect to $t$ , the derivative of $1.7 t$ with respect to $t$ is $1.7 \frac{d}{d t} [t]$ .

$1.7 \frac{d}{d t} [t] + \frac{d}{d t} [0.3]$

Step 6.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$1.7 \cdot 1 + \frac{d}{d t} [0.3]$

Step 6.1.3.3

Multiply $1.7$ by $1$ .

$1.7 + \frac{d}{d t} [0.3]$

Step 6.1.4

Differentiate using the Constant Rule.

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Step 6.1.4.1

Since $0.3$ is constant with respect to $t$ , the derivative of $0.3$ with respect to $t$ is $0$ .

$1.7 + 0$

Step 6.1.4.2

Add $1.7$ and $0$ .

$1.7$

Step 6.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$19.21 \int \sin (u_{1}) \frac{1}{1.7} d u_{1} + \int - 16.32 \cos (1.7 t + 0.3) d t$

Step 7

Combine $\sin (u_{1})$ and $\frac{1}{1.7}$ .

$19.21 \int \frac{\sin (u_{1})}{1.7} d u_{1} + \int - 16.32 \cos (1.7 t + 0.3) d t$

Step 8

Since $\frac{1}{1.7}$ is constant with respect to $u_{1}$ , move $\frac{1}{1.7}$ out of the integral.

$19.21 (\frac{1}{1.7} \int \sin (u_{1}) d u_{1}) + \int - 16.32 \cos (1.7 t + 0.3) d t$

Step 9

Combine $\frac{1}{1.7}$ and $19.21$ .

$\frac{19.21}{1.7} \int \sin (u_{1}) d u_{1} + \int - 16.32 \cos (1.7 t + 0.3) d t$

Step 10

The integral of $\sin (u_{1})$ with respect to $u_{1}$ is $- \cos (u_{1})$ .

$\frac{19.21}{1.7} (- \cos (u_{1}) + C) + \int - 16.32 \cos (1.7 t + 0.3) d t$

Step 11

Since $- 16.32$ is constant with respect to $t$ , move $- 16.32$ out of the integral.

$\frac{19.21}{1.7} (- \cos (u_{1}) + C) - 16.32 \int \cos (1.7 t + 0.3) d t$

Step 12

Let $u_{2} = 1.7 t + 0.3$ . Then $d u_{2} = 1.7 d t$ , so $\frac{1}{1.7} d u_{2} = d t$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 12.1

Let $u_{2} = 1.7 t + 0.3$ . Find $\frac{d u_{2}}{d t}$ .

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Step 12.1.1

Differentiate $1.7 t + 0.3$ .

$\frac{d}{d t} [1.7 t + 0.3]$

Step 12.1.2

By the Sum Rule, the derivative of $1.7 t + 0.3$ with respect to $t$ is $\frac{d}{d t} [1.7 t] + \frac{d}{d t} [0.3]$ .

$\frac{d}{d t} [1.7 t] + \frac{d}{d t} [0.3]$

Step 12.1.3

Evaluate $\frac{d}{d t} [1.7 t]$ .

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Step 12.1.3.1

Since $1.7$ is constant with respect to $t$ , the derivative of $1.7 t$ with respect to $t$ is $1.7 \frac{d}{d t} [t]$ .

$1.7 \frac{d}{d t} [t] + \frac{d}{d t} [0.3]$

Step 12.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$1.7 \cdot 1 + \frac{d}{d t} [0.3]$

Step 12.1.3.3

Multiply $1.7$ by $1$ .

$1.7 + \frac{d}{d t} [0.3]$

Step 12.1.4

Differentiate using the Constant Rule.

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Step 12.1.4.1

Since $0.3$ is constant with respect to $t$ , the derivative of $0.3$ with respect to $t$ is $0$ .

$1.7 + 0$

Step 12.1.4.2

Add $1.7$ and $0$ .

$1.7$

Step 12.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{19.21}{1.7} (- \cos (u_{1}) + C) - 16.32 \int \cos (u_{2}) \frac{1}{1.7} d u_{2}$

Step 13

Combine $\cos (u_{2})$ and $\frac{1}{1.7}$ .

$\frac{19.21}{1.7} (- \cos (u_{1}) + C) - 16.32 \int \frac{\cos (u_{2})}{1.7} d u_{2}$

Step 14

Since $\frac{1}{1.7}$ is constant with respect to $u_{2}$ , move $\frac{1}{1.7}$ out of the integral.

$\frac{19.21}{1.7} (- \cos (u_{1}) + C) - 16.32 (\frac{1}{1.7} \int \cos (u_{2}) d u_{2})$

Step 15

Simplify.

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Step 15.1

Combine $\frac{1}{1.7}$ and $- 16.32$ .

$\frac{19.21}{1.7} (- \cos (u_{1}) + C) + \frac{- 16.32}{1.7} \int \cos (u_{2}) d u_{2}$

Step 15.2

Move the negative in front of the fraction.

$\frac{19.21}{1.7} (- \cos (u_{1}) + C) - \frac{16.32}{1.7} \int \cos (u_{2}) d u_{2}$

Step 16

The integral of $\cos (u_{2})$ with respect to $u_{2}$ is $\sin (u_{2})$ .

$\frac{19.21}{1.7} (- \cos (u_{1}) + C) - \frac{16.32}{1.7} (\sin (u_{2}) + C)$

Step 17

Simplify.

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Step 17.1

Simplify.

$11.3 (- \cos (u_{1})) - \frac{16.32}{1.7} \sin (u_{2}) + C$

Step 17.2

Multiply $- 1$ by $11.3$ .

$- 11.3 \cos (u_{1}) - \frac{16.32}{1.7} \sin (u_{2}) + C$

Step 18

Substitute back in for each integration substitution variable.

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Step 18.1

Replace all occurrences of $u_{1}$ with $1.7 t + 0.3$ .

$- 11.3 \cos (1.7 t + 0.3) - \frac{16.32}{1.7} \sin (u_{2}) + C$

Step 18.2

Replace all occurrences of $u_{2}$ with $1.7 t + 0.3$ .

$- 11.3 \cos (1.7 t + 0.3) - \frac{16.32}{1.7} \sin (1.7 t + 0.3) + C$

Step 19

Simplify.

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Step 19.1

Divide $16.32$ by $1.7$ .

$- 11.3 \cos (1.7 t + 0.3) - 1 \cdot 9.6 \sin (1.7 t + 0.3) + C$

Step 19.2

Multiply $- 1$ by $9.6$ .

$- 11.3 \cos (1.7 t + 0.3) - 9.6 \sin (1.7 t + 0.3) + C$

Step 20

The answer is the antiderivative of the function $f (t) = 19.21 \sin (1.7 t + 0.3) - 16.32 \cos (1.7 t + 0.3)$ .

$F (t) =$ $- 11.3 \cos (1.7 t + 0.3) - 9.6 \sin (1.7 t + 0.3) + C$