Calculus Examples

$\sin (8 x)$

Step 1

Write $\sin (8 x)$ as a function.

$f (x) = \sin (8 x)$

Step 2

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 3

Set up the integral to solve.

$F (x) = \int \sin (8 x) d x$

Step 4

Let $u = 8 x$ . Then $d u = 8 d x$ , so $\frac{1}{8} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 4.1

Let $u = 8 x$ . Find $\frac{d u}{d x}$ .

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Step 4.1.1

Differentiate $8 x$ .

$\frac{d}{d x} [8 x]$

Step 4.1.2

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$8 \frac{d}{d x} [x]$

Step 4.1.3

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

$8 \cdot 1$

Step 4.1.4

Multiply $8$ by $1$ .

$8$

Step 4.2

Rewrite the problem using $u$ and $d u$ .

$\int \sin (u) \frac{1}{8} d u$

Step 5

Combine $\sin (u)$ and $\frac{1}{8}$ .

$\int \frac{\sin (u)}{8} d u$

Step 6

Since $\frac{1}{8}$ is constant with respect to $u$ , move $\frac{1}{8}$ out of the integral.

$\frac{1}{8} \int \sin (u) d u$

Step 7

The integral of $\sin (u)$ with respect to $u$ is $- \cos (u)$ .

$\frac{1}{8} (- \cos (u) + C)$

Step 8

Simplify.

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Step 8.1

Simplify.

$\frac{1}{8} (- \cos (u)) + C$

Step 8.2

Combine $\frac{1}{8}$ and $\cos (u)$ .

$- \frac{\cos (u)}{8} + C$

Step 9

Replace all occurrences of $u$ with $8 x$ .

$- \frac{\cos (8 x)}{8} + C$

Step 10

Reorder terms.

$- \frac{1}{8} \cos (8 x) + C$

Step 11

The answer is the antiderivative of the function $f (x) = \sin (8 x)$ .

$F (x) =$ $- \frac{1}{8} \cos (8 x) + C$