Calculus Examples

$\ln (\sec (x))$

Step 1

Find the asymptotes.

Tap for more steps...

Step 1.1

For any $y = \sec (x)$ , vertical asymptotes occur at $x = \frac{π}{2} + n π$ , where $n$ is an integer. Use the basic period for $y = s e c (x)$ , $(- \frac{π}{2}, \frac{3 π}{2})$ , to find the vertical asymptotes for $y = \ln (\sec (x))$ . Set the inside of the secant function, $b x + c$ , for $y = a \sec (b x + c) + d$ equal to $- \frac{π}{2}$ to find where the vertical asymptote occurs for $y = \ln (\sec (x))$ .

$\sec (x) = - \frac{π}{2}$

Step 1.2

Solve for $x$ .

Tap for more steps...

Step 1.2.1

Take the inverse secant of both sides of the equation to extract $x$ from inside the secant.

$x = arcsec (- \frac{π}{2})$

Step 1.2.2

Simplify the right side.

Tap for more steps...

Step 1.2.2.1

Evaluate $arcsec (- \frac{π}{2})$ .

$x = 2.26090341$

Step 1.2.3

The secant function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 (3.14159265) - 2.26090341$

Step 1.2.4

Solve for $x$ .

Tap for more steps...

Step 1.2.4.1

Remove parentheses.

$x = 2 (3.14159265) - 2.26090341$

Step 1.2.4.2

Simplify $2 (3.14159265) - 2.26090341$ .

Tap for more steps...

Step 1.2.4.2.1

Multiply $2$ by $3.14159265$ .

$x = 6.2831853 - 2.26090341$

Step 1.2.4.2.2

Subtract $2.26090341$ from $6.2831853$ .

$x = 4.02228188$

Step 1.2.5

Find the period of $\sec (x)$ .

Tap for more steps...

Step 1.2.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 1.2.5.3

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

$\frac{2 π}{1}$

Step 1.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.6

The period of the $\sec (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 2.26090341 + 2 π n, 4.02228188 + 2 π n$ , for any integer $n$

Step 1.3

Set the inside of the secant function $\sec (x)$ equal to $\frac{3 π}{2}$ .

$\sec (x) = \frac{3 π}{2}$

Step 1.4

Solve for $x$ .

Tap for more steps...

Step 1.4.1

Take the inverse secant of both sides of the equation to extract $x$ from inside the secant.

$x = arcsec (\frac{3 π}{2})$

Step 1.4.2

Simplify the right side.

Tap for more steps...

Step 1.4.2.1

Evaluate $arcsec (\frac{3 π}{2})$ .

$x = 1.3569639$

Step 1.4.3

The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 (3.14159265) - 1.3569639$

Step 1.4.4

Solve for $x$ .

Tap for more steps...

Step 1.4.4.1

Remove parentheses.

$x = 2 (3.14159265) - 1.3569639$

Step 1.4.4.2

Simplify $2 (3.14159265) - 1.3569639$ .

Tap for more steps...

Step 1.4.4.2.1

Multiply $2$ by $3.14159265$ .

$x = 6.2831853 - 1.3569639$

Step 1.4.4.2.2

Subtract $1.3569639$ from $6.2831853$ .

$x = 4.9262214$

Step 1.4.5

Find the period of $\sec (x)$ .

Tap for more steps...

Step 1.4.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.4.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 1.4.5.3

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

$\frac{2 π}{1}$

Step 1.4.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.4.6

The period of the $\sec (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 1.3569639 + 2 π n, 4.9262214 + 2 π n$ , for any integer $n$

Step 1.5

The basic period for $y = \ln (\sec (x))$ will occur at $(2.26090341 + 2 π n, 4.02228188 + 2 π n, 1.3569639 + 2 π n, 4.9262214 + 2 π n)$ , where $2.26090341 + 2 π n, 4.02228188 + 2 π n$ and $1.3569639 + 2 π n, 4.9262214 + 2 π n$ are vertical asymptotes.

$(2.26090341 + 2 π n, 4.02228188 + 2 π n, 1.3569639 + 2 π n, 4.9262214 + 2 π n)$

Step 1.6

Find the period $\frac{2 π}{| b |}$ to find where the vertical asymptotes exist. Vertical asymptotes occur every half period.

Tap for more steps...

Step 1.6.1

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

$\frac{2 π}{1}$

Step 1.6.2

Divide $2 π$ by $1$ .

$2 π$

Step 1.7

The vertical asymptotes for $y = \ln (\sec (x))$ occur at $2.26090341 + 2 π n, 4.02228188 + 2 π n$ , $1.3569639 + 2 π n, 4.9262214 + 2 π n$ , and every $π n$ , where $n$ is an integer. This is half of the period.

$π n$

Step 1.8

There are only vertical asymptotes for secant and cosecant functions.

Vertical Asymptotes: $x = 2.26090341 + 2 π n, 4.02228188 + 2 π n + π n$ for any integer $n$

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 2.26090341 + 2 π n, 4.02228188 + 2 π n + π n$ for any integer $n$

No Horizontal Asymptotes

No Oblique Asymptotes

Step 2

Find the point at $x = 1$ .

Tap for more steps...

Step 2.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \ln (\sec (1))$

Step 2.2

Simplify the result.

Tap for more steps...

Step 2.2.1

Evaluate $\sec (1)$ .

$f (1) = \ln (1.85081571)$

Step 2.2.2

The final answer is $\ln (1.85081571)$ .

$\ln (1.85081571)$

Step 3

Find the point at $x = 5$ .

Tap for more steps...

Step 3.1

Replace the variable $x$ with $5$ in the expression.

$f (5) = \ln (\sec (5))$

Step 3.2

Simplify the result.

Tap for more steps...

Step 3.2.1

Evaluate $\sec (5)$ .

$f (5) = \ln (3.52532008)$

Step 3.2.2

The final answer is $\ln (3.52532008)$ .

$\ln (3.52532008)$

Step 4

Find the point at $x = 6$ .

Tap for more steps...

Step 4.1

Replace the variable $x$ with $6$ in the expression.

$f (6) = \ln (\sec (6))$

Step 4.2

Simplify the result.

Tap for more steps...

Step 4.2.1

Evaluate $\sec (6)$ .

$f (6) = \ln (1.04148192)$

Step 4.2.2

The final answer is $\ln (1.04148192)$ .

$\ln (1.04148192)$

Step 5

The log function can be graphed using the vertical asymptote at $x = 2.26090341 + 2 π n, 4.02228188 + 2 π n + π n (f o r) (a n y) (i n t e g e r) n$ and the points $(1, 0.61562647), (5, 1.25997123), (6, 0.04064462)$ .

Vertical Asymptote: $x = 2.26090341 + 2 π n, 4.02228188 + 2 π n + π n (f o r) (a n y) (i n t e g e r) n$

$\begin{array}{c} x & y \\ 1 & 0.616 \\ 5 & 1.26 \\ 6 & 0.041 \end{array}$

Step 6