Calculus Examples

$f (x) = 5 \sin (\frac{x}{5}) + \cos (2 x)$

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.

$F (x) = \int 5 \sin (\frac{x}{5}) + \cos (2 x) d x$

Step 3

Split the single integral into multiple integrals.

$\int 5 \sin (\frac{x}{5}) d x + \int \cos (2 x) d x$

Step 4

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$5 \int \sin (\frac{x}{5}) d x + \int \cos (2 x) d x$

Step 5

Let $u_{1} = \frac{x}{5}$ . Then $d u_{1} = \frac{1}{5} d x$ , so $5 d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 5.1

Let $u_{1} = \frac{x}{5}$ . Find $\frac{d u_{1}}{d x}$ .

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Step 5.1.1

Differentiate $\frac{x}{5}$ .

$\frac{d}{d x} [\frac{x}{5}]$

Step 5.1.2

Since $\frac{1}{5}$ is constant with respect to $x$ , the derivative of $\frac{x}{5}$ with respect to $x$ is $\frac{1}{5} \frac{d}{d x} [x]$ .

$\frac{1}{5} \frac{d}{d x} [x]$

Step 5.1.3

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

$\frac{1}{5} \cdot 1$

Step 5.1.4

Multiply $\frac{1}{5}$ by $1$ .

$\frac{1}{5}$

Step 5.2

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

$5 \int \sin (u_{1}) \frac{1}{\frac{1}{5}} d u_{1} + \int \cos (2 x) d x$

Step 6

Simplify.

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Step 6.1

Multiply by the reciprocal of the fraction to divide by $\frac{1}{5}$ .

$5 \int \sin (u_{1}) (1 \cdot 5) d u_{1} + \int \cos (2 x) d x$

Step 6.2

Multiply $5$ by $1$ .

$5 \int \sin (u_{1}) \cdot 5 d u_{1} + \int \cos (2 x) d x$

Step 6.3

Move $5$ to the left of $\sin (u_{1})$ .

$5 \int 5 \sin (u_{1}) d u_{1} + \int \cos (2 x) d x$

Step 7

Since $5$ is constant with respect to $u_{1}$ , move $5$ out of the integral.

$5 (5 \int \sin (u_{1}) d u_{1}) + \int \cos (2 x) d x$

Step 8

Multiply $5$ by $5$ .

$25 \int \sin (u_{1}) d u_{1} + \int \cos (2 x) d x$

Step 9

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

$25 (- \cos (u_{1}) + C) + \int \cos (2 x) d x$

Step 10

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

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Step 10.1

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

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Step 10.1.1

Differentiate $2 x$ .

$\frac{d}{d x} [2 x]$

Step 10.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x]$

Step 10.1.3

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

$2 \cdot 1$

Step 10.1.4

Multiply $2$ by $1$ .

$2$

Step 10.2

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

$25 (- \cos (u_{1}) + C) + \int \cos (u_{2}) \frac{1}{2} d u_{2}$

Step 11

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

$25 (- \cos (u_{1}) + C) + \int \frac{\cos (u_{2})}{2} d u_{2}$

Step 12

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

$25 (- \cos (u_{1}) + C) + \frac{1}{2} \int \cos (u_{2}) d u_{2}$

Step 13

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

$25 (- \cos (u_{1}) + C) + \frac{1}{2} (\sin (u_{2}) + C)$

Step 14

Simplify.

$- 25 \cos (u_{1}) + \frac{1}{2} \sin (u_{2}) + C$

Step 15

Substitute back in for each integration substitution variable.

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Step 15.1

Replace all occurrences of $u_{1}$ with $\frac{x}{5}$ .

$- 25 \cos (\frac{x}{5}) + \frac{1}{2} \sin (u_{2}) + C$

Step 15.2

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

$- 25 \cos (\frac{x}{5}) + \frac{1}{2} \sin (2 x) + C$

Step 16

Reorder terms.

$- 25 \cos (\frac{1}{5} x) + \frac{1}{2} \sin (2 x) + C$

Step 17

The answer is the antiderivative of the function $f (x) = 5 \sin (\frac{x}{5}) + \cos (2 x)$ .

$F (x) =$ $- 25 \cos (\frac{1}{5} x) + \frac{1}{2} \sin (2 x) + C$