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

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

Simplify.

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

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

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

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

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

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

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Simplify.

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Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

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Rewrite.

$\frac{4}{1}$

Divide $4$ by $1$ .

$4$

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

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

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

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

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

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

Simplify.

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

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

Multiply $4$ by $4$ .

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

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

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

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

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

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

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

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