Calculus Examples

$\int_{0}^{\frac{π}{12}} 1 - \cos (2 x) d x$

Step 1

Split the single integral into multiple integrals.

$\int_{0}^{\frac{π}{12}} d x + \int_{0}^{\frac{π}{12}} - \cos (2 x) d x$

Step 2

Apply the constant rule.

$x]_{0}^{\frac{π}{12}} + \int_{0}^{\frac{π}{12}} - \cos (2 x) d x$

Step 3

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

$x]_{0}^{\frac{π}{12}} - \int_{0}^{\frac{π}{12}} \cos (2 x) d x$

Step 4

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 4.1

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

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Step 4.1.1

Differentiate $2 x$ .

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

Step 4.1.2

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 4.1.3

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 4.1.4

Multiply $2$ by $1$ .

$2$

Step 4.2

Substitute the lower limit in for $x$ in $u = 2 x$ .

$u_{lower} = 2 \cdot 0$

Step 4.3

Multiply $2$ by $0$ .

$u_{lower} = 0$

Step 4.4

Substitute the upper limit in for $x$ in $u = 2 x$ .

$u_{upper} = 2 \frac{π}{12}$

Step 4.5

Cancel the common factor of $2$ .

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Step 4.5.1

Factor $2$ out of $12$ .

$u_{upper} = 2 \frac{π}{2 (6)}$

Step 4.5.2

Cancel the common factor.

$u_{upper} = 2 \frac{π}{2 \cdot 6}$

Step 4.5.3

Rewrite the expression.

$u_{upper} = \frac{π}{6}$

Step 4.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = \frac{π}{6}$

Step 4.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$x]_{0}^{\frac{π}{12}} - \int_{0}^{\frac{π}{6}} \cos (u) \frac{1}{2} d u$

Step 5

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

$x]_{0}^{\frac{π}{12}} - \int_{0}^{\frac{π}{6}} \frac{\cos (u)}{2} d u$

Step 6

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

$x]_{0}^{\frac{π}{12}} - (\frac{1}{2} \int_{0}^{\frac{π}{6}} \cos (u) d u)$

Step 7

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

$x]_{0}^{\frac{π}{12}} - \frac{1}{2} \sin (u)]_{0}^{\frac{π}{6}}$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $x$ at $\frac{π}{12}$ and at $0$ .

$(\frac{π}{12}) + 0 - \frac{1}{2} \sin (u)]_{0}^{\frac{π}{6}}$

Step 8.2

Evaluate $\sin (u)$ at $\frac{π}{6}$ and at $0$ .

$(\frac{π}{12}) + 0 - \frac{1}{2} ((\sin (\frac{π}{6})) - \sin (0))$

Step 8.3

Add $\frac{π}{12}$ and $0$ .

$\frac{π}{12} - \frac{1}{2} (\sin (\frac{π}{6}) - \sin (0))$

Step 9

Simplify.

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Step 9.1

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$\frac{π}{12} - \frac{1}{2} (\frac{1}{2} - \sin (0))$

Step 9.2

The exact value of $\sin (0)$ is $0$ .

$\frac{π}{12} - \frac{1}{2} (\frac{1}{2} - 0)$

Step 9.3

Multiply $- 1$ by $0$ .

$\frac{π}{12} - \frac{1}{2} (\frac{1}{2} + 0)$

Step 9.4

Add $\frac{1}{2}$ and $0$ .

$\frac{π}{12} - \frac{1}{2} \cdot \frac{1}{2}$

Step 9.5

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

$\frac{π}{12} - \frac{1}{2 \cdot 2}$

Step 9.6

Multiply $2$ by $2$ .

$\frac{π}{12} - \frac{1}{4}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$\frac{π}{12} - \frac{1}{4}$

Decimal Form:

$0.01179938 \dots$