Calculus Examples

$\int_{- \frac{π}{3}}^{\frac{π}{3}} [(- \sec (x)) - (- 2)] d x$

Step 1

Remove parentheses.

$\int_{- \frac{π}{3}}^{\frac{π}{3}} (- \sec (x)) - (- 2) d x$

Step 2

Multiply $- 1$ by $- 2$ .

$\int_{- \frac{π}{3}}^{\frac{π}{3}} - \sec (x) + 2 d x$

Step 3

Split the single integral into multiple integrals.

$\int_{- \frac{π}{3}}^{\frac{π}{3}} - \sec (x) d x + \int_{- \frac{π}{3}}^{\frac{π}{3}} 2 d x$

Step 4

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

$- \int_{- \frac{π}{3}}^{\frac{π}{3}} \sec (x) d x + \int_{- \frac{π}{3}}^{\frac{π}{3}} 2 d x$

Step 5

The integral of $\sec (x)$ with respect to $x$ is $\ln (| \sec (x) + \tan (x) |)$ .

$- (\ln (| \sec (x) + \tan (x) |)]_{- \frac{π}{3}}^{\frac{π}{3}}) + \int_{- \frac{π}{3}}^{\frac{π}{3}} 2 d x$

Step 6

Apply the constant rule.

$- (\ln (| \sec (x) + \tan (x) |)]_{- \frac{π}{3}}^{\frac{π}{3}}) + 2 x]_{- \frac{π}{3}}^{\frac{π}{3}}$

Step 7

Simplify the answer.

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Step 7.1

Substitute and simplify.

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Step 7.1.1

Evaluate $\ln (| \sec (x) + \tan (x) |)$ at $\frac{π}{3}$ and at $- \frac{π}{3}$ .

$- ((\ln (| \sec (\frac{π}{3}) + \tan (\frac{π}{3}) |)) - \ln (| \sec (- \frac{π}{3}) + \tan (- \frac{π}{3}) |)) + 2 x]_{- \frac{π}{3}}^{\frac{π}{3}}$

Step 7.1.2

Evaluate $2 x$ at $\frac{π}{3}$ and at $- \frac{π}{3}$ .

$- ((\ln (| \sec (\frac{π}{3}) + \tan (\frac{π}{3}) |)) - \ln (| \sec (- \frac{π}{3}) + \tan (- \frac{π}{3}) |)) + (2 \frac{π}{3}) - 2 (- \frac{π}{3})$

Step 7.1.3

Simplify.

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Step 7.1.3.1

Combine $2$ and $\frac{π}{3}$ .

$- (\ln (| \sec (\frac{π}{3}) + \tan (\frac{π}{3}) |) - \ln (| \sec (- \frac{π}{3}) + \tan (- \frac{π}{3}) |)) + \frac{2 π}{3} - 2 (- \frac{π}{3})$

Step 7.1.3.2

Multiply $- 1$ by $- 2$ .

$- (\ln (| \sec (\frac{π}{3}) + \tan (\frac{π}{3}) |) - \ln (| \sec (- \frac{π}{3}) + \tan (- \frac{π}{3}) |)) + \frac{2 π}{3} + 2 \frac{π}{3}$

Step 7.1.3.3

Combine $2$ and $\frac{π}{3}$ .

$- (\ln (| \sec (\frac{π}{3}) + \tan (\frac{π}{3}) |) - \ln (| \sec (- \frac{π}{3}) + \tan (- \frac{π}{3}) |)) + \frac{2 π}{3} + \frac{2 π}{3}$

Step 7.1.3.4

Combine the numerators over the common denominator.

$- (\ln (| \sec (\frac{π}{3}) + \tan (\frac{π}{3}) |) - \ln (| \sec (- \frac{π}{3}) + \tan (- \frac{π}{3}) |)) + \frac{2 π + 2 π}{3}$

Step 7.1.3.5

Add $2 π$ and $2 π$ .

$- (\ln (| \sec (\frac{π}{3}) + \tan (\frac{π}{3}) |) - \ln (| \sec (- \frac{π}{3}) + \tan (- \frac{π}{3}) |)) + \frac{4 π}{3}$

Step 7.2

Simplify.

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Step 7.2.1

The exact value of $\sec (\frac{π}{3})$ is $2$ .

$- (\ln (| 2 + \tan (\frac{π}{3}) |) - \ln (| \sec (- \frac{π}{3}) + \tan (- \frac{π}{3}) |)) + \frac{4 π}{3}$

Step 7.2.2

The exact value of $\tan (\frac{π}{3})$ is $\sqrt{3}$ .

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

Step 7.2.3

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

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

Step 7.3

Simplify.

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Step 7.3.1

$2 + \sqrt{3}$ is approximately $3.7320508$ which is positive so remove the absolute value

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

Step 7.3.2

Simplify the denominator.

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Step 7.3.2.1

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$- \ln (\frac{2 + \sqrt{3}}{| \sec (\frac{5 π}{3}) + \tan (- \frac{π}{3}) |}) + \frac{4 π}{3}$

Step 7.3.2.2

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

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

Step 7.3.2.3

The exact value of $\sec (\frac{π}{3})$ is $2$ .

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

Step 7.3.2.4

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$- \ln (\frac{2 + \sqrt{3}}{| 2 + \tan (\frac{5 π}{3}) |}) + \frac{4 π}{3}$

Step 7.3.2.5

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

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

Step 7.3.2.6

The exact value of $\tan (\frac{π}{3})$ is $\sqrt{3}$ .

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

Step 7.3.2.7

$2 - \sqrt{3}$ is approximately $0.26794919$ which is positive so remove the absolute value

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

Step 7.3.3

To write $- \ln (\frac{2 + \sqrt{3}}{2 - \sqrt{3}})$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \ln (\frac{2 + \sqrt{3}}{2 - \sqrt{3}}) \cdot \frac{3}{3} + \frac{4 π}{3}$

Step 7.3.4

Combine $- \ln (\frac{2 + \sqrt{3}}{2 - \sqrt{3}})$ and $\frac{3}{3}$ .

$\frac{- \ln (\frac{2 + \sqrt{3}}{2 - \sqrt{3}}) \cdot 3}{3} + \frac{4 π}{3}$

Step 7.3.5

Combine the numerators over the common denominator.

$\frac{- \ln (\frac{2 + \sqrt{3}}{2 - \sqrt{3}}) \cdot 3 + 4 π}{3}$

Step 7.3.6

Multiply $3$ by $- 1$ .

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

Step 8

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$1.55487441 \dots$