Question

What is this equal how can I solve similar trigonometric integrals like this one

Answer 1

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