Calculus Examples

\int_{\frac{π}{4}}^{\frac{3 π}{4}} csc (x) d x

$\int_{\frac{π}{4}}^{\frac{3 π}{4}} \csc (x) d x$

Step 1

This integral could not be completed using u-substitution. Mathway will use another method.

Step 2

The integral of

csc (x)

$\csc (x)$ with respect to

x

$x$ is

ln (| csc (x) - cot (x) |)

$\ln (| \csc (x) - \cot (x) |)$ .

ln (| csc (x) - cot (x) |)]_{\frac{π}{4}}^{\frac{3 π}{4}}

$\ln (| \csc (x) - \cot (x) |)]_{\frac{π}{4}}^{\frac{3 π}{4}}$

Step 3

Simplify the answer.

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Step 3.1

Evaluate

ln (| csc (x) - cot (x) |)

$\ln (| \csc (x) - \cot (x) |)$ at

\frac{3 π}{4}

$\frac{3 π}{4}$ and at

\frac{π}{4}

$\frac{π}{4}$ .

ln (∣ ∣ ∣ csc (\frac{3 π}{4}) - cot (\frac{3 π}{4}) ∣ ∣ ∣) - ln (∣ ∣ csc (\frac{π}{4}) - cot (\frac{π}{4}) ∣ ∣)

$\ln (| \csc (\frac{3 π}{4}) - \cot (\frac{3 π}{4}) |) - \ln (| \csc (\frac{π}{4}) - \cot (\frac{π}{4}) |)$

Step 3.2

Simplify.

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Step 3.2.1

The exact value of

csc (\frac{π}{4})

$\csc (\frac{π}{4})$ is

\sqrt{2}

$\sqrt{2}$ .

ln (∣ ∣ ∣ csc (\frac{3 π}{4}) - cot (\frac{3 π}{4}) ∣ ∣ ∣) - ln (∣ ∣ \sqrt{2} - cot (\frac{π}{4}) ∣ ∣)

$\ln (| \csc (\frac{3 π}{4}) - \cot (\frac{3 π}{4}) |) - \ln (| \sqrt{2} - \cot (\frac{π}{4}) |)$

Step 3.2.2

The exact value of

cot (\frac{π}{4})

$\cot (\frac{π}{4})$ is

1

$1$ .

ln (∣ ∣ ∣ csc (\frac{3 π}{4}) - cot (\frac{3 π}{4}) ∣ ∣ ∣) - ln (∣ ∣ \sqrt{2} - 1 \cdot 1 ∣ ∣)

$\ln (| \csc (\frac{3 π}{4}) - \cot (\frac{3 π}{4}) |) - \ln (| \sqrt{2} - 1 \cdot 1 |)$

Step 3.2.3

Multiply

- 1

$- 1$ by

1

$1$ .

ln (∣ ∣ ∣ csc (\frac{3 π}{4}) - cot (\frac{3 π}{4}) ∣ ∣ ∣) - ln (∣ ∣ \sqrt{2} - 1 ∣ ∣)

$\ln (| \csc (\frac{3 π}{4}) - \cot (\frac{3 π}{4}) |) - \ln (| \sqrt{2} - 1 |)$

Step 3.2.4

Use the quotient property of logarithms,

{log}_{b} (x) - {log}_{b} (y) = {log}_{b} (\frac{x}{y})

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

ln ⎛ ⎜ ⎝ \frac{∣ ∣ csc (\frac{3 π}{4}) - cot (\frac{3 π}{4}) ∣ ∣}{∣ ∣ \sqrt{2} - 1 ∣ ∣} ⎞ ⎟ ⎠

$\ln (\frac{| \csc (\frac{3 π}{4}) - \cot (\frac{3 π}{4}) |}{| \sqrt{2} - 1 |})$

ln ⎛ ⎜ ⎝ \frac{∣ ∣ csc (\frac{3 π}{4}) - cot (\frac{3 π}{4}) ∣ ∣}{∣ ∣ \sqrt{2} - 1 ∣ ∣} ⎞ ⎟ ⎠

$\ln (\frac{| \csc (\frac{3 π}{4}) - \cot (\frac{3 π}{4}) |}{| \sqrt{2} - 1 |})$

Step 3.3

Simplify.

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Step 3.3.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cotangent is negative in the second quadrant.

ln ⎛ ⎜ ⎝ \frac{∣ ∣ csc (\frac{3 π}{4}) - - cot (\frac{π}{4}) ∣ ∣}{∣ ∣ \sqrt{2} - 1 ∣ ∣} ⎞ ⎟ ⎠

$\ln (\frac{| \csc (\frac{3 π}{4}) - - \cot (\frac{π}{4}) |}{| \sqrt{2} - 1 |})$

Step 3.3.2

The exact value of

cot (\frac{π}{4})

$\cot (\frac{π}{4})$ is

1

$1$ .

ln ⎛ ⎜ ⎝ \frac{∣ ∣ csc (\frac{3 π}{4}) - (- 1 \cdot 1) ∣ ∣}{∣ ∣ \sqrt{2} - 1 ∣ ∣} ⎞ ⎟ ⎠

$\ln (\frac{| \csc (\frac{3 π}{4}) - (- 1 \cdot 1) |}{| \sqrt{2} - 1 |})$

Step 3.3.3

Multiply

- 1

$- 1$ by

1

$1$ .

ln ⎛ ⎜ ⎝ \frac{∣ ∣ csc (\frac{3 π}{4}) - - 1 ∣ ∣}{∣ ∣ \sqrt{2} - 1 ∣ ∣} ⎞ ⎟ ⎠

$\ln (\frac{| \csc (\frac{3 π}{4}) - - 1 |}{| \sqrt{2} - 1 |})$

Step 3.3.4

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

ln ⎛ ⎜ ⎝ \frac{∣ ∣ csc (\frac{3 π}{4}) + 1 ∣ ∣}{∣ ∣ \sqrt{2} - 1 ∣ ∣} ⎞ ⎟ ⎠

$\ln (\frac{| \csc (\frac{3 π}{4}) + 1 |}{| \sqrt{2} - 1 |})$

Step 3.3.5

Simplify the numerator.

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Step 3.3.5.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

ln (\frac{∣ ∣ csc (\frac{π}{4}) + 1 ∣ ∣}{∣ ∣ \sqrt{2} - 1 ∣ ∣})

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

Step 3.3.5.2

The exact value of

csc (\frac{π}{4})

$\csc (\frac{π}{4})$ is

\sqrt{2}

$\sqrt{2}$ .

ln (\frac{∣ ∣ \sqrt{2} + 1 ∣ ∣}{∣ ∣ \sqrt{2} - 1 ∣ ∣})

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

Step 3.3.5.3

\sqrt{2} + 1

$\sqrt{2} + 1$ is approximately

2.41421356

$2.41421356$ which is positive so remove the absolute value

ln (\frac{\sqrt{2} + 1}{∣ ∣ \sqrt{2} - 1 ∣ ∣})

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

ln (\frac{\sqrt{2} + 1}{∣ ∣ \sqrt{2} - 1 ∣ ∣})

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

Step 3.3.6

\sqrt{2} - 1

$\sqrt{2} - 1$ is approximately

0.41421356

$0.41421356$ which is positive so remove the absolute value

ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})

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

ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})

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

ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})

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

Step 4

Simplify.

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Step 4.1

Multiply

\frac{\sqrt{2} + 1}{\sqrt{2} - 1}

$\frac{\sqrt{2} + 1}{\sqrt{2} - 1}$ by

\frac{\sqrt{2} + 1}{\sqrt{2} + 1}

$\frac{\sqrt{2} + 1}{\sqrt{2} + 1}$ .

ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1})

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

Step 4.2

Multiply

\frac{\sqrt{2} + 1}{\sqrt{2} - 1}

$\frac{\sqrt{2} + 1}{\sqrt{2} - 1}$ by

\frac{\sqrt{2} + 1}{\sqrt{2} + 1}

$\frac{\sqrt{2} + 1}{\sqrt{2} + 1}$ .

ln ⎛ ⎜ ⎝ \frac{(\sqrt{2} + 1) (\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)} ⎞ ⎟ ⎠

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

Step 4.3

Expand the denominator using the FOIL method.

ln ⎛ ⎜ ⎝ \frac{(\sqrt{2} + 1) (\sqrt{2} + 1)}{{\sqrt{2}}^{2} + \sqrt{2} - \sqrt{2} - 1} ⎞ ⎟ ⎠

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

Step 4.4

Simplify.

ln ⎛ ⎜ ⎝ \frac{(\sqrt{2} + 1) (\sqrt{2} + 1)}{1} ⎞ ⎟ ⎠

$\ln (\frac{(\sqrt{2} + 1) (\sqrt{2} + 1)}{1})$

Step 4.5

Divide

(\sqrt{2} + 1) (\sqrt{2} + 1)

$(\sqrt{2} + 1) (\sqrt{2} + 1)$ by

1

$1$ .

ln ((\sqrt{2} + 1) (\sqrt{2} + 1))

$\ln ((\sqrt{2} + 1) (\sqrt{2} + 1))$

Step 4.6

Rewrite

ln ((\sqrt{2} + 1) (\sqrt{2} + 1))

$\ln ((\sqrt{2} + 1) (\sqrt{2} + 1))$ as

ln (\sqrt{2} + 1) + ln (\sqrt{2} + 1)

$\ln (\sqrt{2} + 1) + \ln (\sqrt{2} + 1)$ .

$\ln (\sqrt{2} + 1) + \ln (\sqrt{2} + 1)$

Step 4.7

Add

$\ln (\sqrt{2} + 1)$ and

$\ln (\sqrt{2} + 1)$ .

$2 \ln (\sqrt{2} + 1)$

Step 5

The result can be shown in multiple forms.

Exact Form:

$2 \ln (\sqrt{2} + 1)$

Decimal Form:

$1.76274717 \dots$