# What is the integration of secx 1

You must mean, either, (1) sec2 x = sec x + 2, or (2) sec(2x) = sec x + 2. Let's first assume that you mean: sec2 x = sec x + 2; whence, if we let s = sec x, we have, s2 = s + 2, s2 - s - 2 = (s - 2)(s + 1) = 0, and s = 2 or -1; that is, sec x = 2 or -1. As, by definition, cos x = 1/sec x , this means that cos x = ½ or -1. Therefore, providing that the first assumption is correct, x = 60°, 180°, or 300°; or, if you prefer, x = ⅓ π, π, or 1⅔ π. Now, let's assume, instead, that you mean: sec(2x) = sec x + 2; whence, 1/(cos(2x) = (1/cos x) + 2. If we let c = cos x, then we have the standard identity, 2c2 - 1 = cos (2x); and, thus, it follows that 1/(cos(2x) = 1/(2c2 - 1) = (1/c) + 2 = (1 + 2c)/c. This gives, 1/(2c2 - 1) = (1 + 2c)/c; (2c2 - 1)(2c + 1) = 4c3 + 2c2 - 2c - 1 = c; and 4c3 + 2c2 - 3c - 1 = (c + 1)(4c2 - 2c - 1) = 0. As our concern is only with real roots, c = cos x = -1; and, therefore, providing that the second assumption is correct, x = 180°; or, if you like, x = π.

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