Graph natural log of sec(x)+tan(x)

Problem

ln(sec(x)+tan(x))

Solution

1. Identify the function to be graphed, which is ƒ(x)=ln(sec(x)+tan(x)) This function is the result of the integral (∫_^)(sec(x)*d(x))

2. Determine the domain by ensuring the argument of the natural logarithm is positive, sec(x)+tan(x)>0 and that sec(x) and tan(x) are defined. This requires x≠π/2+n*π for any integer n Specifically, for the principal branch, the function is defined where cos(x)>0 and (1+sin(x))/cos(x)>0 which occurs on intervals like (−π/2,π/2)

3. Analyze symmetry and periodicity. Since sec(x) and tan(x) have a period of 2*π the function ƒ(x) also has a period of 2*π However, because of the logarithm and the signs of the trigonometric functions, the function is typically analyzed on its primary domain.

4. Find vertical asymptotes where the argument approaches zero or where the trigonometric functions are undefined. As x→π/2− sec(x)+tan(x)→∞ so ƒ(x)→∞ As x→−π/2+ sec(x)+tan(x)→0 so ƒ(x)→−∞

5. Determine the derivative to find the slope. The derivative is ƒ(x)′=sec(x) On the interval (−π/2,π/2) sec(x)>0 meaning the function is strictly increasing.

6. Identify the y-intercept by evaluating ƒ(0) Since sec(0)=1 and tan(0)=0 ƒ(0)=ln(1)=0

7. Sketch the curve starting from the bottom-left at the vertical asymptote x=−π/2 passing through the origin (0,0) with a slope of 1 and rising toward the vertical asymptote at x=π/2

Final Answer

y=ln(sec(x)+tan(x))

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