Calculus Examples

$\int_{\frac{π}{8}}^{\frac{π}{4}} 8 \csc (2 x) - 8 \cot (2 x) d x$

Step 1

Split the single integral into multiple integrals.

$\int_{\frac{π}{8}}^{\frac{π}{4}} 8 \csc (2 x) d x + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 2

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$8 \int_{\frac{π}{8}}^{\frac{π}{4}} \csc (2 x) d x + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 3

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

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Step 3.1

Let $u_{1} = 2 x$ . Find $\frac{d u_{1}}{d x}$ .

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Step 3.1.1

Differentiate $2 x$ .

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

Step 3.1.2

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 3.1.3

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 3.1.4

Multiply $2$ by $1$ .

$2$

Step 3.2

Substitute the lower limit in for $x$ in $u_{1} = 2 x$ .

$u_{lower} = 2 \frac{π}{8}$

Step 3.3

Cancel the common factor of $2$ .

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Step 3.3.1

Factor $2$ out of $8$ .

$u_{lower} = 2 \frac{π}{2 (4)}$

Step 3.3.2

Cancel the common factor.

$u_{lower} = 2 \frac{π}{2 \cdot 4}$

Step 3.3.3

Rewrite the expression.

$u_{lower} = \frac{π}{4}$

Step 3.4

Substitute the upper limit in for $x$ in $u_{1} = 2 x$ .

$u_{upper} = 2 \frac{π}{4}$

Step 3.5

Cancel the common factor of $2$ .

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Step 3.5.1

Factor $2$ out of $4$ .

$u_{upper} = 2 \frac{π}{2 (2)}$

Step 3.5.2

Cancel the common factor.

$u_{upper} = 2 \frac{π}{2 \cdot 2}$

Step 3.5.3

Rewrite the expression.

$u_{upper} = \frac{π}{2}$

Step 3.6

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

$u_{lower} = \frac{π}{4}$

$u_{upper} = \frac{π}{2}$

Step 3.7

Rewrite the problem using $u_{1}$ , $d u_{1}$ , and the new limits of integration.

$8 \int_{\frac{π}{4}}^{\frac{π}{2}} \csc (u_{1}) \frac{1}{2} d u_{1} + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 4

Combine $\csc (u_{1})$ and $\frac{1}{2}$ .

$8 \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\csc (u_{1})}{2} d u_{1} + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 5

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

$8 (\frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \csc (u_{1}) d u_{1}) + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 6

Simplify.

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Step 6.1

Combine $\frac{1}{2}$ and $8$ .

$\frac{8}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \csc (u_{1}) d u_{1} + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 6.2

Cancel the common factor of $8$ and $2$ .

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Step 6.2.1

Factor $2$ out of $8$ .

$\frac{2 \cdot 4}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \csc (u_{1}) d u_{1} + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 6.2.2

Cancel the common factors.

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Step 6.2.2.1

Factor $2$ out of $2$ .

$\frac{2 \cdot 4}{2 (1)} \int_{\frac{π}{4}}^{\frac{π}{2}} \csc (u_{1}) d u_{1} + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 6.2.2.2

Cancel the common factor.

$\frac{2 \cdot 4}{2 \cdot 1} \int_{\frac{π}{4}}^{\frac{π}{2}} \csc (u_{1}) d u_{1} + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 6.2.2.3

Rewrite the expression.

$\frac{4}{1} \int_{\frac{π}{4}}^{\frac{π}{2}} \csc (u_{1}) d u_{1} + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 6.2.2.4

Divide $4$ by $1$ .

$4 \int_{\frac{π}{4}}^{\frac{π}{2}} \csc (u_{1}) d u_{1} + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 7

The integral of $\csc (u_{1})$ with respect to $u_{1}$ is $\ln (| \csc (u_{1}) - \cot (u_{1}) |)$ .

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) + \int_{\frac{π}{8}}^{\frac{π}{4}} - 8 \cot (2 x) d x$

Step 8

Since $- 8$ is constant with respect to $x$ , move $- 8$ out of the integral.

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) - 8 \int_{\frac{π}{8}}^{\frac{π}{4}} \cot (2 x) d x$

Step 9

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

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Step 9.1

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

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Step 9.1.1

Differentiate $2 x$ .

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

Step 9.1.2

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 9.1.3

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 9.1.4

Multiply $2$ by $1$ .

$2$

Step 9.2

Substitute the lower limit in for $x$ in $u_{2} = 2 x$ .

$u_{lower} = 2 \frac{π}{8}$

Step 9.3

Cancel the common factor of $2$ .

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Step 9.3.1

Factor $2$ out of $8$ .

$u_{lower} = 2 \frac{π}{2 (4)}$

Step 9.3.2

Cancel the common factor.

$u_{lower} = 2 \frac{π}{2 \cdot 4}$

Step 9.3.3

Rewrite the expression.

$u_{lower} = \frac{π}{4}$

Step 9.4

Substitute the upper limit in for $x$ in $u_{2} = 2 x$ .

$u_{upper} = 2 \frac{π}{4}$

Step 9.5

Cancel the common factor of $2$ .

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Step 9.5.1

Factor $2$ out of $4$ .

$u_{upper} = 2 \frac{π}{2 (2)}$

Step 9.5.2

Cancel the common factor.

$u_{upper} = 2 \frac{π}{2 \cdot 2}$

Step 9.5.3

Rewrite the expression.

$u_{upper} = \frac{π}{2}$

Step 9.6

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

$u_{lower} = \frac{π}{4}$

$u_{upper} = \frac{π}{2}$

Step 9.7

Rewrite the problem using $u_{2}$ , $d u_{2}$ , and the new limits of integration.

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) - 8 \int_{\frac{π}{4}}^{\frac{π}{2}} \cot (u_{2}) \frac{1}{2} d u_{2}$

Step 10

Combine $\cot (u_{2})$ and $\frac{1}{2}$ .

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) - 8 \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\cot (u_{2})}{2} d u_{2}$

Step 11

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

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) - 8 (\frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \cot (u_{2}) d u_{2})$

Step 12

Simplify.

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Step 12.1

Combine $\frac{1}{2}$ and $- 8$ .

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) + \frac{- 8}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \cot (u_{2}) d u_{2}$

Step 12.2

Cancel the common factor of $- 8$ and $2$ .

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Step 12.2.1

Factor $2$ out of $- 8$ .

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) + \frac{2 \cdot - 4}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \cot (u_{2}) d u_{2}$

Step 12.2.2

Cancel the common factors.

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Step 12.2.2.1

Factor $2$ out of $2$ .

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) + \frac{2 \cdot - 4}{2 (1)} \int_{\frac{π}{4}}^{\frac{π}{2}} \cot (u_{2}) d u_{2}$

Step 12.2.2.2

Cancel the common factor.

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) + \frac{2 \cdot - 4}{2 \cdot 1} \int_{\frac{π}{4}}^{\frac{π}{2}} \cot (u_{2}) d u_{2}$

Step 12.2.2.3

Rewrite the expression.

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) + \frac{- 4}{1} \int_{\frac{π}{4}}^{\frac{π}{2}} \cot (u_{2}) d u_{2}$

Step 12.2.2.4

Divide $- 4$ by $1$ .

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) - 4 \int_{\frac{π}{4}}^{\frac{π}{2}} \cot (u_{2}) d u_{2}$

Step 13

The integral of $\cot (u_{2})$ with respect to $u_{2}$ is $\ln (| \sin (u_{2}) |)$ .

$4 (\ln (| \csc (u_{1}) - \cot (u_{1}) |)]_{\frac{π}{4}}^{\frac{π}{2}}) - 4 (\ln (| \sin (u_{2}) |)]_{\frac{π}{4}}^{\frac{π}{2}})$

Step 14

Substitute and simplify.

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Step 14.1

Evaluate $\ln (| \csc (u_{1}) - \cot (u_{1}) |)$ at $\frac{π}{2}$ and at $\frac{π}{4}$ .

$4 ((\ln (| \csc (\frac{π}{2}) - \cot (\frac{π}{2}) |)) - \ln (| \csc (\frac{π}{4}) - \cot (\frac{π}{4}) |)) - 4 (\ln (| \sin (u_{2}) |)]_{\frac{π}{4}}^{\frac{π}{2}})$

Step 14.2

Evaluate $\ln (| \sin (u_{2}) |)$ at $\frac{π}{2}$ and at $\frac{π}{4}$ .

$4 ((\ln (| \csc (\frac{π}{2}) - \cot (\frac{π}{2}) |)) - \ln (| \csc (\frac{π}{4}) - \cot (\frac{π}{4}) |)) - 4 ((\ln (| \sin (\frac{π}{2}) |)) - \ln (| \sin (\frac{π}{4}) |))$

Step 14.3

Remove parentheses.

$4 (\ln (| \csc (\frac{π}{2}) - \cot (\frac{π}{2}) |) - \ln (| \csc (\frac{π}{4}) - \cot (\frac{π}{4}) |)) - 4 (\ln (| \sin (\frac{π}{2}) |) - \ln (| \sin (\frac{π}{4}) |))$

Step 15

Simplify.

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Step 15.1

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

$4 (\ln (| 1 - \cot (\frac{π}{2}) |) - \ln (| \csc (\frac{π}{4}) - \cot (\frac{π}{4}) |)) - 4 (\ln (| \sin (\frac{π}{2}) |) - \ln (| \sin (\frac{π}{4}) |))$

Step 15.2

The exact value of $\cot (\frac{π}{2})$ is $0$ .

$4 (\ln (| 1 - 0 |) - \ln (| \csc (\frac{π}{4}) - \cot (\frac{π}{4}) |)) - 4 (\ln (| \sin (\frac{π}{2}) |) - \ln (| \sin (\frac{π}{4}) |))$

Step 15.3

The exact value of $\csc (\frac{π}{4})$ is $\sqrt{2}$ .

$4 (\ln (| 1 - 0 |) - \ln (| \sqrt{2} - \cot (\frac{π}{4}) |)) - 4 (\ln (| \sin (\frac{π}{2}) |) - \ln (| \sin (\frac{π}{4}) |))$

Step 15.4

The exact value of $\cot (\frac{π}{4})$ is $1$ .

$4 (\ln (| 1 - 0 |) - \ln (| \sqrt{2} - 1 \cdot 1 |)) - 4 (\ln (| \sin (\frac{π}{2}) |) - \ln (| \sin (\frac{π}{4}) |))$

Step 15.5

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

$4 (\ln (| 1 - 0 |) - \ln (| \sqrt{2} - 1 \cdot 1 |)) - 4 (\ln (| 1 |) - \ln (| \sin (\frac{π}{4}) |))$

Step 15.6

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$4 (\ln (| 1 - 0 |) - \ln (| \sqrt{2} - 1 \cdot 1 |)) - 4 (\ln (| 1 |) - \ln (| \frac{\sqrt{2}}{2} |))$

Step 15.7

Multiply $- 1$ by $0$ .

$4 (\ln (| 1 + 0 |) - \ln (| \sqrt{2} - 1 \cdot 1 |)) - 4 (\ln (| 1 |) - \ln (| \frac{\sqrt{2}}{2} |))$

Step 15.8

Add $1$ and $0$ .

$4 (\ln (| 1 |) - \ln (| \sqrt{2} - 1 \cdot 1 |)) - 4 (\ln (| 1 |) - \ln (| \frac{\sqrt{2}}{2} |))$

Step 15.9

Multiply $- 1$ by $1$ .

$4 (\ln (| 1 |) - \ln (| \sqrt{2} - 1 |)) - 4 (\ln (| 1 |) - \ln (| \frac{\sqrt{2}}{2} |))$

Step 15.10

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$4 \ln (\frac{| 1 |}{| \sqrt{2} - 1 |}) - 4 (\ln (| 1 |) - \ln (| \frac{\sqrt{2}}{2} |))$

Step 15.11

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$4 \ln (\frac{| 1 |}{| \sqrt{2} - 1 |}) - 4 \ln (\frac{| 1 |}{| \frac{\sqrt{2}}{2} |})$

Step 16

Simplify.

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Step 16.1

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$4 \ln (\frac{1}{| \sqrt{2} - 1 |}) - 4 \ln (\frac{| 1 |}{| \frac{\sqrt{2}}{2} |})$

Step 16.2

$\sqrt{2} - 1$ is approximately $0.41421356$ which is positive so remove the absolute value

$4 \ln (\frac{1}{\sqrt{2} - 1}) - 4 \ln (\frac{| 1 |}{| \frac{\sqrt{2}}{2} |})$

Step 16.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$4 \ln (\frac{1}{\sqrt{2} - 1}) - 4 \ln (\frac{1}{| \frac{\sqrt{2}}{2} |})$

Step 16.4

$\frac{\sqrt{2}}{2}$ is approximately $0.70710678$ which is positive so remove the absolute value

$4 \ln (\frac{1}{\sqrt{2} - 1}) - 4 \ln (\frac{1}{\frac{\sqrt{2}}{2}})$

Step 16.5

Multiply the numerator by the reciprocal of the denominator.

$4 \ln (\frac{1}{\sqrt{2} - 1}) - 4 \ln (1 \frac{2}{\sqrt{2}})$

Step 16.6

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

$4 \ln (\frac{1}{\sqrt{2} - 1}) - 4 \ln (\frac{2}{\sqrt{2}})$

Step 17

The result can be shown in multiple forms.

Exact Form:

$4 \ln (\frac{1}{\sqrt{2} - 1}) - 4 \ln (\frac{2}{\sqrt{2}})$

Decimal Form:

$2.13919998 \dots$