Calculus Examples

$\int x \cos (3 x) d x$

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = \cos (3 x)$ .

$x (\frac{1}{3} \sin (3 x)) - \int \frac{1}{3} \sin (3 x) d x$

Combine fractions.

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Write $\sin (3 x)$ as a fraction with denominator $1$ .

$x (\frac{1}{3} \cdot \frac{\sin (3 x)}{1}) - \int \frac{1}{3} \sin (3 x) d x$

Multiply $\frac{1}{3}$ and $\frac{\sin (3 x)}{1}$ .

$x \frac{\sin (3 x)}{3} - \int \frac{1}{3} \sin (3 x) d x$

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

$\frac{x}{1} \cdot \frac{\sin (3 x)}{3} - \int \frac{1}{3} \sin (3 x) d x$

Multiply $\frac{x}{1}$ and $\frac{\sin (3 x)}{3}$ .

$\frac{x \sin (3 x)}{3} - \int \frac{1}{3} \sin (3 x) d x$

Write $\sin (3 x)$ as a fraction with denominator $1$ .

$\frac{x \sin (3 x)}{3} - \int \frac{1}{3} \cdot \frac{\sin (3 x)}{1} d x$

Multiply $\frac{1}{3}$ and $\frac{\sin (3 x)}{1}$ .

$\frac{x \sin (3 x)}{3} - \int \frac{\sin (3 x)}{3} d x$

Since $\frac{1}{3}$ is constant with respect to $x$ , move $\frac{1}{3}$ out of the integral.

$\frac{x \sin (3 x)}{3} - (\frac{1}{3} \int \sin (3 x) d x)$

Remove parentheses.

$\frac{x \sin (3 x)}{3} - \frac{1}{3} \int \sin (3 x) d x$

Let $u = 3 x$ . Then $d u = 3 d x$ , so $\frac{1}{3} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

$\frac{x \sin (3 x)}{3} - \frac{1}{3} \int \sin (u) \frac{1}{3} d u$

Combine fractions.

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

$\frac{x \sin (3 x)}{3} - \frac{1}{3} \int \frac{\sin (u)}{1} \cdot \frac{1}{3} d u$

Multiply $\frac{\sin (u)}{1}$ and $\frac{1}{3}$ .

$\frac{x \sin (3 x)}{3} - \frac{1}{3} \int \frac{\sin (u)}{3} d u$

Since $\frac{1}{3}$ is constant with respect to $u$ , move $\frac{1}{3}$ out of the integral.

$\frac{x \sin (3 x)}{3} - \frac{1}{3} (\frac{1}{3} \int \sin (u) d u)$

Combine fractions.

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Multiply $\frac{1}{3}$ and $\frac{1}{3}$ .

$\frac{x \sin (3 x)}{3} - \frac{1}{3 \cdot 3} \int \sin (u) d u$

Multiply $3$ by $3$ .

$\frac{x \sin (3 x)}{3} - \frac{1}{9} \int \sin (u) d u$

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

$\frac{x \sin (3 x)}{3} - \frac{1}{9} (- \cos (u) + C)$

Simplify the answer.

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Rewrite $\frac{x \sin (3 x)}{3} - \frac{1}{9} (- \cos (u) + C)$ as $\frac{x \sin (3 x)}{3} + \frac{\cos (u)}{9} + C$ .

$\frac{x \sin (3 x)}{3} + \frac{\cos (u)}{9} + C$

Replace all occurrences of $u$ with $3 x$ .

$\frac{x \sin (3 x)}{3} + \frac{\cos (3 x)}{9} + C$

Reorder terms.

$\frac{1}{3} x \sin (3 x) + \frac{1}{9} \cos (3 x) + C$

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