Answer:
ln|sec θ + tan θ| + C
Step-by-step explanation:
The integrals of basic trig functions are:
∫ sin θ dθ = -cos θ + C
∫ cos θ dθ = sin θ + C
∫ csc θ dθ = -ln|csc θ + cot θ| + C
∫ sec θ dθ = ln|sec θ + tan θ| + C
∫ tan θ dθ = -ln|cos θ| + C
∫ cot θ dθ = ln|sin θ| + C
The integral of sec θ can be proven by multiplying and dividing by sec θ + tan θ, then using ∫ du/u = ln|u| + C.
∫ sec θ dθ
∫ sec θ (sec θ + tan θ) / (sec θ + tan θ) dθ
∫ (sec² θ + sec θ tan θ) / (sec θ + tan θ) dθ
ln|sec θ + tan θ| + C
