Calculus Examples

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

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

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

Combine fractions.

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Write $\cos (4 x)$ as a fraction with denominator $1$ .

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

Multiply $\frac{\cos (4 x)}{1}$ and $\frac{1}{4}$ to get $\frac{\cos (4 x)}{4}$ .

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

Write $x$ as a fraction with denominator $1$ .

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

Multiply $\frac{x}{1}$ and $\frac{\cos (4 x)}{4}$ to get $\frac{x \cos (4 x)}{4}$ .

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

Write $\cos (4 x)$ as a fraction with denominator $1$ .

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

Multiply $\frac{\cos (4 x)}{1}$ and $\frac{1}{4}$ to get $\frac{\cos (4 x)}{4}$ .

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

Since $- 1$ is constant with respect to $x$ , the integral of $- \frac{\cos (4 x)}{4}$ with respect to $x$ is $- \int \frac{\cos (4 x)}{4} d x$ .

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

Multiply.

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Multiply $- 1$ by $- 1$ to get $1$ .

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

Multiply $\int \frac{\cos (4 x)}{4} d x$ by $1$ to get $\int \frac{\cos (4 x)}{4} d x$ .

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

Since $\frac{1}{4}$ is constant with respect to $x$ , the integral of $\frac{\cos (4 x)}{4}$ with respect to $x$ is $\frac{1}{4} \int \cos (4 x) d x$ .

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

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

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

Combine fractions.

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Write $\cos (u)$ as a fraction with denominator $1$ .

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

Multiply $\frac{\cos (u)}{1}$ and $\frac{1}{4}$ to get $\frac{\cos (u)}{4}$ .

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

Since $\frac{1}{4}$ is constant with respect to $u$ , the integral of $\frac{\cos (u)}{4}$ with respect to $u$ is $\frac{1}{4} \int \cos (u) d u$ .

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

Combine fractions.

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Multiply $\frac{1}{4}$ and $\frac{1}{4}$ to get $\frac{1}{4 \cdot 4}$ .

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

Multiply $4$ by $4$ to get $16$ .

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

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

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

Simplify the answer.

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

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

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

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

Reorder terms.

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

Remove parentheses around $\frac{1}{4} x \cos (4 x)$ .

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

Remove parentheses around $\frac{1}{4} x$ .

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

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

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