Calculus Examples

$f (x) = \frac{1}{3} \cos (6 x) - 4 \sin (4 x)$

Step 1

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

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

Step 2

Set up the integral to solve.

$F (x) = \int \frac{1}{3} \cdot \cos (6 x) - 4 \sin (4 x) d x$

Step 3

Split the single integral into multiple integrals.

$\int \frac{1}{3} \cdot \cos (6 x) d x + \int - 4 \sin (4 x) d x$

Step 4

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int \cos (6 x) d x + \int - 4 \sin (4 x) d x$

Step 5

Let $u_{1} = 6 x$ . Then $d u_{1} = 6 d x$ , so $\frac{1}{6} d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 5.1

Let $u_{1} = 6 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 5.1.1

Differentiate $6 x$ .

$\frac{d}{d x} [6 x]$

Step 5.1.2

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$6 \frac{d}{d x} [x]$

Step 5.1.3

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

$6 \cdot 1$

Step 5.1.4

Multiply $6$ by $1$ .

$6$

Step 5.2

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

$\frac{1}{3} \int \cos (u_{1}) \frac{1}{6} d u_{1} + \int - 4 \sin (4 x) d x$

Step 6

Combine $\cos (u_{1})$ and $\frac{1}{6}$ .

$\frac{1}{3} \int \frac{\cos (u_{1})}{6} d u_{1} + \int - 4 \sin (4 x) d x$

Step 7

Since $\frac{1}{6}$ is constant with respect to $u_{1}$ , move $\frac{1}{6}$ out of the integral.

$\frac{1}{3} (\frac{1}{6} \int \cos (u_{1}) d u_{1}) + \int - 4 \sin (4 x) d x$

Step 8

Simplify.

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Step 8.1

Multiply $\frac{1}{6}$ by $\frac{1}{3}$ .

$\frac{1}{6 \cdot 3} \int \cos (u_{1}) d u_{1} + \int - 4 \sin (4 x) d x$

Step 8.2

Multiply $6$ by $3$ .

$\frac{1}{18} \int \cos (u_{1}) d u_{1} + \int - 4 \sin (4 x) d x$

Step 9

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

$\frac{1}{18} (\sin (u_{1}) + C) + \int - 4 \sin (4 x) d x$

Step 10

Since $- 4$ is constant with respect to $x$ , move $- 4$ out of the integral.

$\frac{1}{18} (\sin (u_{1}) + C) - 4 \int \sin (4 x) d x$

Step 11

Let $u_{2} = 4 x$ . Then $d u_{2} = 4 d x$ , so $\frac{1}{4} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 11.1

Let $u_{2} = 4 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 11.1.1

Differentiate $4 x$ .

$\frac{d}{d x} [4 x]$

Step 11.1.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x]$

Step 11.1.3

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

$4 \cdot 1$

Step 11.1.4

Multiply $4$ by $1$ .

$4$

Step 11.2

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

$\frac{1}{18} (\sin (u_{1}) + C) - 4 \int \sin (u_{2}) \frac{1}{4} d u_{2}$

Step 12

Combine $\sin (u_{2})$ and $\frac{1}{4}$ .

$\frac{1}{18} (\sin (u_{1}) + C) - 4 \int \frac{\sin (u_{2})}{4} d u_{2}$

Step 13

Since $\frac{1}{4}$ is constant with respect to $u_{2}$ , move $\frac{1}{4}$ out of the integral.

$\frac{1}{18} (\sin (u_{1}) + C) - 4 (\frac{1}{4} \int \sin (u_{2}) d u_{2})$

Step 14

Simplify.

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Step 14.1

Combine $\frac{1}{4}$ and $- 4$ .

$\frac{1}{18} (\sin (u_{1}) + C) + \frac{- 4}{4} \int \sin (u_{2}) d u_{2}$

Step 14.2

Cancel the common factor of $- 4$ and $4$ .

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Step 14.2.1

Factor $4$ out of $- 4$ .

$\frac{1}{18} (\sin (u_{1}) + C) + \frac{4 \cdot - 1}{4} \int \sin (u_{2}) d u_{2}$

Step 14.2.2

Cancel the common factors.

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Step 14.2.2.1

Factor $4$ out of $4$ .

$\frac{1}{18} (\sin (u_{1}) + C) + \frac{4 \cdot - 1}{4 (1)} \int \sin (u_{2}) d u_{2}$

Step 14.2.2.2

Cancel the common factor.

$\frac{1}{18} (\sin (u_{1}) + C) + \frac{4 \cdot - 1}{4 \cdot 1} \int \sin (u_{2}) d u_{2}$

Step 14.2.2.3

Rewrite the expression.

$\frac{1}{18} (\sin (u_{1}) + C) + \frac{- 1}{1} \int \sin (u_{2}) d u_{2}$

Step 14.2.2.4

Divide $- 1$ by $1$ .

$\frac{1}{18} (\sin (u_{1}) + C) - \int \sin (u_{2}) d u_{2}$

Step 15

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

$\frac{1}{18} (\sin (u_{1}) + C) - (- \cos (u_{2}) + C)$

Step 16

Simplify.

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Step 16.1

Simplify.

$\frac{1}{18} \sin (u_{1}) - - \cos (u_{2}) + C$

Step 16.2

Simplify.

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Step 16.2.1

Multiply $- 1$ by $- 1$ .

$\frac{1}{18} \sin (u_{1}) + 1 \cos (u_{2}) + C$

Step 16.2.2

Multiply $\cos (u_{2})$ by $1$ .

$\frac{1}{18} \sin (u_{1}) + \cos (u_{2}) + C$

Step 17

Substitute back in for each integration substitution variable.

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Step 17.1

Replace all occurrences of $u_{1}$ with $6 x$ .

$\frac{1}{18} \sin (6 x) + \cos (u_{2}) + C$

Step 17.2

Replace all occurrences of $u_{2}$ with $4 x$ .

$\frac{1}{18} \sin (6 x) + \cos (4 x) + C$

Step 18

The answer is the antiderivative of the function $f (x) = \frac{1}{3} \cdot \cos (6 x) - 4 \sin (4 x)$ .

$F (x) =$ $\frac{1}{18} \sin (6 x) + \cos (4 x) + C$