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How do you integrate ∫sec²θ - 1 dθ?

a) ∫sec²θ - 1 dθ = tanθ - θ + C
b) ∫sec²θ - 1 dθ = tanθ - θ
c) ∫sec²θ - 1 dθ = tanθ - 1
d) ∫sec²θ - 1 dθ = θ - tanθ + C

User Stackflow
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Final answer:

To integrate ∫(sec²θ - 1) dθ, we can rewrite sec²θ - 1 as tan²θ. Then, using trigonometric identities, we integrate both terms separately and add the results together to get the final answer: ∫(sec²θ - 1) dθ = secθ - ln|cosθ| + C.

Step-by-step explanation:

To integrate ∫(sec²θ - 1) dθ, we can rewrite sec²θ - 1 as tan²θ.

Then, using the trigonometric identity tan²θ + 1 = sec²θ, we have:

∫(tan²θ) dθ = ∫(tanθ)(tanθ) dθ = ∫tanθ secθ dθ - ∫tanθ dθ

Now we can integrate both terms separately:

∫tanθ secθ dθ = secθ + C1

∫tanθ dθ = -ln|cosθ| + C2

Adding the two results together, we get the final answer: ∫(sec²θ - 1) dθ = secθ - ln|cosθ| + C

User Marc Alexander
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