passpadi

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 + c
passpadi
©2023 Passpadi. All rights reserved.