2009
Mathematics
67ca00ce0c643c71d77dee1f
Evaluate ∫sec2θ dθ?
A.
sec θ tan θ + k
B.
tan θ + k
C.
2sec θ + k
D.
sec θ + k
Correct Answer: tan θ + k
Explanation
∫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
∫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 + cTHIS WEEK's
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