Calculus Examples

$f (x) = 3 \cos (\frac{x}{3}) + \sin (3 x)$

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.

$F (x) = \int 3 \cos (\frac{x}{3}) + \sin (3 x) d x$

Step 3

Split the single integral into multiple integrals.

$\int 3 \cos (\frac{x}{3}) d x + \int \sin (3 x) d x$

Step 4

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

$3 \int \cos (\frac{x}{3}) d x + \int \sin (3 x) d x$

Step 5

Let $u_{1} = \frac{x}{3}$ . Then $d u_{1} = \frac{1}{3} d x$ , so $3 d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 5.1

Let $u_{1} = \frac{x}{3}$ . Find $\frac{d u_{1}}{d x}$ .

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Step 5.1.1

Differentiate $\frac{x}{3}$ .

$\frac{d}{d x} [\frac{x}{3}]$

Step 5.1.2

Since $\frac{1}{3}$ is constant with respect to $x$ , the derivative of $\frac{x}{3}$ with respect to $x$ is $\frac{1}{3} \frac{d}{d x} [x]$ .

$\frac{1}{3} \frac{d}{d x} [x]$

Step 5.1.3

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

$\frac{1}{3} \cdot 1$

Step 5.1.4

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

$\frac{1}{3}$

Step 5.2

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

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

Step 6

Simplify.

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Step 6.1

Multiply by the reciprocal of the fraction to divide by $\frac{1}{3}$ .

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

Step 6.2

Multiply $3$ by $1$ .

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

Step 6.3

Move $3$ to the left of $\cos (u_{1})$ .

$3 \int 3 \cos (u_{1}) d u_{1} + \int \sin (3 x) d x$

Step 7

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

$3 (3 \int \cos (u_{1}) d u_{1}) + \int \sin (3 x) d x$

Step 8

Multiply $3$ by $3$ .

$9 \int \cos (u_{1}) d u_{1} + \int \sin (3 x) d x$

Step 9

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

$9 (\sin (u_{1}) + C) + \int \sin (3 x) d x$

Step 10

Let $u_{2} = 3 x$ . Then $d u_{2} = 3 d x$ , so $\frac{1}{3} d u_{2} = d x$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 10.1

Let $u_{2} = 3 x$ . Find $\frac{d u_{2}}{d x}$ .

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Step 10.1.1

Differentiate $3 x$ .

$\frac{d}{d x} [3 x]$

Step 10.1.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$3 \frac{d}{d x} [x]$

Step 10.1.3

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

$3 \cdot 1$

Step 10.1.4

Multiply $3$ by $1$ .

$3$

Step 10.2

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

$9 (\sin (u_{1}) + C) + \int \sin (u_{2}) \frac{1}{3} d u_{2}$

Step 11

Combine $\sin (u_{2})$ and $\frac{1}{3}$ .

$9 (\sin (u_{1}) + C) + \int \frac{\sin (u_{2})}{3} d u_{2}$

Step 12

Since $\frac{1}{3}$ is constant with respect to $u_{2}$ , move $\frac{1}{3}$ out of the integral.

$9 (\sin (u_{1}) + C) + \frac{1}{3} \int \sin (u_{2}) d u_{2}$

Step 13

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

$9 (\sin (u_{1}) + C) + \frac{1}{3} (- \cos (u_{2}) + C)$

Step 14

Simplify.

$9 \sin (u_{1}) - \frac{\cos (u_{2})}{3} + C$

Step 15

Substitute back in for each integration substitution variable.

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Step 15.1

Replace all occurrences of $u_{1}$ with $\frac{x}{3}$ .

$9 \sin (\frac{x}{3}) - \frac{\cos (u_{2})}{3} + C$

Step 15.2

Replace all occurrences of $u_{2}$ with $3 x$ .

$9 \sin (\frac{x}{3}) - \frac{\cos (3 x)}{3} + C$

Step 16

Reorder terms.

$9 \sin (\frac{1}{3} x) - \frac{1}{3} \cos (3 x) + C$

Step 17

The answer is the antiderivative of the function $f (x) = 3 \cos (\frac{x}{3}) + \sin (3 x)$ .

$F (x) =$ $9 \sin (\frac{1}{3} x) - \frac{1}{3} \cos (3 x) + C$