Calculus Examples

cos (8 x)

$\cos (8 x)$

Step 1

Let

u = 8 x

$u = 8 x$ . Then

d u = 8 d x

$d u = 8 d x$ , so

\frac{1}{8} d u = d x

$\frac{1}{8} d u = d x$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

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Step 1.1

Let

u = 8 x

$u = 8 x$ . Find

\frac{d u}{d x}

$\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate

8 x

$8 x$ .

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

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

Step 1.1.2

Since

8

$8$ is constant with respect to

x

$x$ , the derivative of

8 x

$8 x$ with respect to

x

$x$ is

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

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

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

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

Step 1.1.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

8 \cdot 1

$8 \cdot 1$

Step 1.1.4

Multiply

8

$8$ by

1

$1$ .

8

$8$

8

$8$

Step 1.2

Rewrite the problem using

u

$u$ and

d u

$d u$ .

\int cos (u) \frac{1}{8} d u

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

\int cos (u) \frac{1}{8} d u

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

Step 2

Combine

cos (u)

$\cos (u)$ and

\frac{1}{8}

$\frac{1}{8}$ .

\int \frac{cos (u)}{8} d u

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

Step 3

Since

\frac{1}{8}

$\frac{1}{8}$ is constant with respect to

u

$u$ , move

\frac{1}{8}

$\frac{1}{8}$ out of the integral.

\frac{1}{8} \int cos (u) d u

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

Step 4

The integral of

cos (u)

$\cos (u)$ with respect to

u

$u$ is

sin (u)

$\sin (u)$ .

\frac{1}{8} (sin (u) + C)

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

Step 5

Simplify.

\frac{1}{8} sin (u) + C

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

Step 6

Replace all occurrences of

u

$u$ with

8 x

$8 x$ .

\frac{1}{8} sin (8 x) + C

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

⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$