Question on: JAMB Mathematics - 2009
Evaluate ∫sec2θ dθ
A
sec θ tan θ + k
B
tan θ + k
C
2sec θ + k
D
sec θ + k
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Correct Option: B
∫sec2θ dθ = ∫ 1/cos2 dθ
∫(cos)-2 dθ,
let u = cos θ
∴∫u-2 = 1/u + c
∫cos θ = sin θ + c
∫sec-2θ = 1/u sin θ + c
= (sinθ / cosθ) + c
= Tan θ + c
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