Calculus Examples

\int x \sin (4 x) d x

$\int x \sin (4 x) d x$

Step 1

Integrate by parts using the formula

\int u d v = u v - \int v d u

$\int u d v = u v - \int v d u$ , where

u = x

$u = x$ and

d v = \sin (4 x)

$d v = \sin (4 x)$ .

x (- \frac{1}{4} \cos (4 x)) - \int - \frac{1}{4} \cos (4 x) d x

$x (- \frac{1}{4} \cos (4 x)) - \int - \frac{1}{4} \cos (4 x) d x$

Step 2

Simplify.

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Step 2.1

Combine

\cos (4 x)

$\cos (4 x)$ and

\frac{1}{4}

$\frac{1}{4}$ .

x (- \frac{\cos (4 x)}{4}) - \int - \frac{1}{4} \cos (4 x) d x

$x (- \frac{\cos (4 x)}{4}) - \int - \frac{1}{4} \cos (4 x) d x$

Step 2.2

Combine

x

$x$ and

\frac{\cos (4 x)}{4}

$\frac{\cos (4 x)}{4}$ .

- \frac{x \cos (4 x)}{4} - \int - \frac{1}{4} \cos (4 x) d x

$- \frac{x \cos (4 x)}{4} - \int - \frac{1}{4} \cos (4 x) d x$

- \frac{x \cos (4 x)}{4} - \int - \frac{1}{4} \cos (4 x) d x

$- \frac{x \cos (4 x)}{4} - \int - \frac{1}{4} \cos (4 x) d x$

Step 3

Since

- \frac{1}{4}

$- \frac{1}{4}$ is constant with respect to

x

$x$ , move

- \frac{1}{4}

$- \frac{1}{4}$ out of the integral.

- \frac{x \cos (4 x)}{4} - (- \frac{1}{4} \int \cos (4 x) d x)

$- \frac{x \cos (4 x)}{4} - (- \frac{1}{4} \int \cos (4 x) d x)$

Step 4

Simplify.

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Step 4.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

- \frac{x \cos (4 x)}{4} + 1 (\frac{1}{4} \int \cos (4 x) d x)

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

Step 4.2

Multiply

\frac{1}{4}

$\frac{1}{4}$ by

1

$1$ .

- \frac{x \cos (4 x)}{4} + \frac{1}{4} \int \cos (4 x) d x

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

- \frac{x \cos (4 x)}{4} + \frac{1}{4} \int \cos (4 x) d x

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

Step 5

Let

u = 4 x

$u = 4 x$ . Then

d u = 4 d x

$d u = 4 d x$ , so

\frac{1}{4} d u = d x

$\frac{1}{4} d u = d x$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

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Step 5.1

Let

u = 4 x

$u = 4 x$ . Find

\frac{d u}{d x}

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

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Step 5.1.1

Differentiate

4 x

$4 x$ .

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

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

Step 5.1.2

Since

4

$4$ is constant with respect to

x

$x$ , the derivative of

4 x

$4 x$ with respect to

x

$x$ is

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

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

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

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

Step 5.1.3

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

$n = 1$ .

4 \cdot 1

$4 \cdot 1$

Step 5.1.4

Multiply

4

$4$ by

1

$1$ .

4

$4$

4

$4$

Step 5.2

Rewrite the problem using

u

$u$ and

d u

$d u$ .

- \frac{x \cos (4 x)}{4} + \frac{1}{4} \int \cos (u) \frac{1}{4} d u

$- \frac{x \cos (4 x)}{4} + \frac{1}{4} \int \cos (u) \frac{1}{4} d u$

- \frac{x \cos (4 x)}{4} + \frac{1}{4} \int \cos (u) \frac{1}{4} d u

$- \frac{x \cos (4 x)}{4} + \frac{1}{4} \int \cos (u) \frac{1}{4} d u$

Step 6

Combine

\cos (u)

$\cos (u)$ and

\frac{1}{4}

$\frac{1}{4}$ .

- \frac{x \cos (4 x)}{4} + \frac{1}{4} \int \frac{\cos (u)}{4} d u

$- \frac{x \cos (4 x)}{4} + \frac{1}{4} \int \frac{\cos (u)}{4} d u$

Step 7

Since

\frac{1}{4}

$\frac{1}{4}$ is constant with respect to

u

$u$ , move

\frac{1}{4}

$\frac{1}{4}$ out of the integral.

- \frac{x \cos (4 x)}{4} + \frac{1}{4} (\frac{1}{4} \int \cos (u) d u)

$- \frac{x \cos (4 x)}{4} + \frac{1}{4} (\frac{1}{4} \int \cos (u) d u)$

Step 8

Simplify.

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Step 8.1

Multiply

\frac{1}{4}

$\frac{1}{4}$ by

\frac{1}{4}

$\frac{1}{4}$ .

- \frac{x \cos (4 x)}{4} + \frac{1}{4 \cdot 4} \int \cos (u) d u

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

Step 8.2

Multiply

4

$4$ by

4

$4$ .

- \frac{x \cos (4 x)}{4} + \frac{1}{16} \int \cos (u) d u

$- \frac{x \cos (4 x)}{4} + \frac{1}{16} \int \cos (u) d u$

- \frac{x \cos (4 x)}{4} + \frac{1}{16} \int \cos (u) d u

$- \frac{x \cos (4 x)}{4} + \frac{1}{16} \int \cos (u) d u$

Step 9

The integral of

\cos (u)

$\cos (u)$ with respect to

u

$u$ is

\sin (u)

$\sin (u)$ .

- \frac{x \cos (4 x)}{4} + \frac{1}{16} (\sin (u) + C)

$- \frac{x \cos (4 x)}{4} + \frac{1}{16} (\sin (u) + C)$

Step 10

Simplify.

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Step 10.1

Rewrite

- \frac{x \cos (4 x)}{4} + \frac{1}{16} (\sin (u) + C)

$- \frac{x \cos (4 x)}{4} + \frac{1}{16} (\sin (u) + C)$ as

- \frac{1}{4} x \cos (4 x) + \frac{1}{16} \sin (u) + C

$- \frac{1}{4} x \cos (4 x) + \frac{1}{16} \sin (u) + C$ .

- \frac{1}{4} x \cos (4 x) + \frac{1}{16} \sin (u) + C

$- \frac{1}{4} x \cos (4 x) + \frac{1}{16} \sin (u) + C$

Step 10.2

Simplify.

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Step 10.2.1

Combine

x

$x$ and

\frac{1}{4}

$\frac{1}{4}$ .

- \frac{x}{4} \cos (4 x) + \frac{1}{16} \sin (u) + C

$- \frac{x}{4} \cos (4 x) + \frac{1}{16} \sin (u) + C$

Step 10.2.2

Combine

\cos (4 x)

$\cos (4 x)$ and

\frac{x}{4}

$\frac{x}{4}$ .

- \frac{\cos (4 x) x}{4} + \frac{1}{16} \sin (u) + C

$- \frac{\cos (4 x) x}{4} + \frac{1}{16} \sin (u) + C$

- \frac{\cos (4 x) x}{4} + \frac{1}{16} \sin (u) + C

$- \frac{\cos (4 x) x}{4} + \frac{1}{16} \sin (u) + C$

- \frac{\cos (4 x) x}{4} + \frac{1}{16} \sin (u) + C

$- \frac{\cos (4 x) x}{4} + \frac{1}{16} \sin (u) + C$

Step 11

Replace all occurrences of

u

$u$ with

4 x

$4 x$ .

- \frac{\cos (4 x) x}{4} + \frac{1}{16} \sin (4 x) + C

$- \frac{\cos (4 x) x}{4} + \frac{1}{16} \sin (4 x) + C$

Step 12

Reorder factors in

\frac{\cos (4 x) x}{4}

$\frac{\cos (4 x) x}{4}$ .

- \frac{1}{4} x \cos (4 x) + \frac{1}{16} \sin (4 x) + C

$- \frac{1}{4} x \cos (4 x) + \frac{1}{16} \sin (4 x) + C$

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