Calculus Examples

$\sin (2 x) d x$

Step 1

Write $\sin (2 x) d x$ as a function.

$f (x) = \sin (2 x) d x$

Step 2

The function $F (x)$ can be found by finding the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Step 3

Set up the integral to solve.

$F (x) = \int \sin (2 x) d x d x$

Step 4

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

$d \int \sin (2 x) x d x$

Step 5

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = \sin (2 x)$ .

$d (x (- \frac{1}{2} \cos (2 x)) - \int - \frac{1}{2} \cos (2 x) d x)$

Step 6

Simplify.

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Step 6.1

Combine $\cos (2 x)$ and $\frac{1}{2}$ .

$d (x (- \frac{\cos (2 x)}{2}) - \int - \frac{1}{2} \cos (2 x) d x)$

Step 6.2

Combine $x$ and $\frac{\cos (2 x)}{2}$ .

$d (- \frac{x \cos (2 x)}{2} - \int - \frac{1}{2} \cos (2 x) d x)$

Step 6.3

Combine $\cos (2 x)$ and $\frac{1}{2}$ .

$d (- \frac{x \cos (2 x)}{2} - \int - \frac{\cos (2 x)}{2} d x)$

Step 7

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

$d (- \frac{x \cos (2 x)}{2} - - \int \frac{\cos (2 x)}{2} d x)$

Step 8

Simplify.

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Step 8.1

Multiply $- 1$ by $- 1$ .

$d (- \frac{x \cos (2 x)}{2} + 1 \int \frac{\cos (2 x)}{2} d x)$

Step 8.2

Multiply $\int \frac{\cos (2 x)}{2} d x$ by $1$ .

$d (- \frac{x \cos (2 x)}{2} + \int \frac{\cos (2 x)}{2} d x)$

Step 9

Since $\frac{1}{2}$ is constant with respect to $x$ , move $\frac{1}{2}$ out of the integral.

$d (- \frac{x \cos (2 x)}{2} + \frac{1}{2} \int \cos (2 x) d x)$

Step 10

Let $u = 2 x$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 10.1

Let $u = 2 x$ . Find $\frac{d u}{d x}$ .

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Step 10.1.1

Differentiate $2 x$ .

$\frac{d}{d x} [2 x]$

Step 10.1.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 \frac{d}{d x} [x]$

Step 10.1.3

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

$2 \cdot 1$

Step 10.1.4

Multiply $2$ by $1$ .

$2$

Step 10.2

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

$d (- \frac{x \cos (2 x)}{2} + \frac{1}{2} \int \cos (u) \frac{1}{2} d u)$

Step 11

Combine $\cos (u)$ and $\frac{1}{2}$ .

$d (- \frac{x \cos (2 x)}{2} + \frac{1}{2} \int \frac{\cos (u)}{2} d u)$

Step 12

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$d (- \frac{x \cos (2 x)}{2} + \frac{1}{2} (\frac{1}{2} \int \cos (u) d u))$

Step 13

Simplify.

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Step 13.1

Multiply $\frac{1}{2}$ by $\frac{1}{2}$ .

$d (- \frac{x \cos (2 x)}{2} + \frac{1}{2 \cdot 2} \int \cos (u) d u)$

Step 13.2

Multiply $2$ by $2$ .

$d (- \frac{x \cos (2 x)}{2} + \frac{1}{4} \int \cos (u) d u)$

Step 14

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

$d (- \frac{x \cos (2 x)}{2} + \frac{1}{4} (\sin (u) + C))$

Step 15

Simplify.

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Step 15.1

Rewrite $d (- \frac{x \cos (2 x)}{2} + \frac{1}{4} (\sin (u) + C))$ as $d (- \frac{1}{2} x \cos (2 x) + \frac{1}{4} \sin (u)) + C$ .

$d (- \frac{1}{2} x \cos (2 x) + \frac{1}{4} \sin (u)) + C$

Step 15.2

Simplify.

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Step 15.2.1

Combine $x$ and $\frac{1}{2}$ .

$d (- \frac{x}{2} \cos (2 x) + \frac{1}{4} \sin (u)) + C$

Step 15.2.2

Combine $\cos (2 x)$ and $\frac{x}{2}$ .

$d (- \frac{\cos (2 x) x}{2} + \frac{1}{4} \sin (u)) + C$

Step 15.2.3

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

$d (- \frac{\cos (2 x) x}{2} + \frac{\sin (u)}{4}) + C$

Step 16

Replace all occurrences of $u$ with $2 x$ .

$d (- \frac{\cos (2 x) x}{2} + \frac{\sin (2 x)}{4}) + C$

Step 17

Simplify.

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Step 17.1

Reorder factors in $\frac{\cos (2 x) x}{2}$ .

$d (- \frac{x \cos (2 x)}{2} + \frac{\sin (2 x)}{4}) + C$

Step 17.2

Reorder terms.

$d (- \frac{1}{2} x \cos (2 x) + \frac{1}{4} \sin (2 x)) + C$

Step 18

The answer is the antiderivative of the function $f (x) = \sin (2 x) d x$ .

$F (x) =$ $d (- \frac{1}{2} x \cos (2 x) + \frac{1}{4} \sin (2 x)) + C$