Calculus Examples

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

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = \sin (4 x)$ .

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

Step 2

Simplify.

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Step 2.1

Combine $\cos (4 x)$ and $\frac{1}{4}$ .

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

Step 2.2

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

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

Step 3

Since $- \frac{1}{4}$ is constant with respect to $x$ , move $- \frac{1}{4}$ out of the integral.

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

Step 4

Simplify.

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Step 4.1

Multiply $- 1$ by $- 1$ .

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

Step 4.2

Multiply $\frac{1}{4}$ by $1$ .

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

Step 5

Let $u = 4 x$ . Then $d u = 4 d x$ , so $\frac{1}{4} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = 4 x$ . Find $\frac{d u}{d x}$ .

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Step 5.1.1

Differentiate $4 x$ .

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

Step 5.1.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

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

Step 5.1.3

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

$4 \cdot 1$

Step 5.1.4

Multiply $4$ by $1$ .

$4$

Step 5.2

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

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

Step 6

Combine $\cos (u)$ and $\frac{1}{4}$ .

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

Step 7

Since $\frac{1}{4}$ is constant with respect to $u$ , move $\frac{1}{4}$ out of the integral.

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

Step 8

Simplify.

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Step 8.1

Multiply $\frac{1}{4}$ by $\frac{1}{4}$ .

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

Step 8.2

Multiply $4$ by $4$ .

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

Step 9

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

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

Step 10

Simplify.

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Step 10.1

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

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

Step 10.2

Simplify.

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Step 10.2.1

Combine $x$ and $\frac{1}{4}$ .

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

Step 10.2.2

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

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

Step 11

Replace all occurrences of $u$ with $4 x$ .

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

Step 12

Reorder factors in $\frac{\cos (4 x) x}{4}$ .

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

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