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I = ∫

**sec**^5(x) dx = ∫ (sin^2(x) + cos^2(x))***sec**^5...**sec**^3(x) dx Note: d(sin(x)***sec**^4(x)) = cos(x)***sec**^4(x) dx + sin(x)*4**sec**...1/2)**sec**(x) dx I = (1/4)sin(x)***sec**^4(x) + (3/4)*(1/2) sin(x)***sec**^2(x) - (3...1 Answers · Science & Mathematics · 30/03/2013

**sec**(2x) = 1/cos(2x) The period of cos (2x) can be found using... In this case B = 2. So the period of cos(2x) = period of**sec**(2x) = 2pi/2 = pi or 180 degrees The amplitude is defined...1 Answers · Science & Mathematics · 15/07/2013

∫

**sec**^4 (x/2) dx = ∫**sec**^2 (x/2) ***sec**^2 (x/2) dx = ∫[1 + tan^2 (x/2)] ***sec**^2 (x/2) dx = ∫**sec**^2 (x... tan(x/2) = u for the**sec**ond integral => (1/2)**sec**^2 (x/2) dx = du =>**sec**^2 (x/2) dx = 2du => Integral...2 Answers · Science & Mathematics · 28/01/2011

First, a few identities: [a]

**sec**x = 1 / cos x [b] csc x = 1 / sin x [c] cos x + n360 = cos x...90. 5) Apply [i], [j] or [k] if possible. So...**sec**-150 = 1/cos -150 [a] = 1/cos 150 [e] = -(1/cos 30) [g] = -(1...1 Answers · Science & Mathematics · 03/12/2007

**sec**(x) doesn't exist when x = pi/2 + pi * k, where k is an integer a / x^2...pi , 0 , pi , 2pi are the values when f(x) doesn't exist f(x) = (**sec**(x) - 1) / x^2 f(x) = (1/cos(x) - 1) / x^2 f(x) = ((1 - cos(x)) / cos(x)) / x^2 f...1 Answers · Science & Mathematics · 03/01/2014

**sec**(t) = -5 First, let's express this as cosine. Note that**sec**(t) = 1/cos(t), so 1/cos(t) = -5. Take the reciprocal of both sides...4 Answers · Science & Mathematics · 12/02/2007

1)

**sec**^2x(1-sin^2x)**sec**^2x ( cos^2x) ----------------- sin^2x + cos^2x = 1 cos^2x/cos^2x ---------------**sec**^2x = 1/cos^2x 1 2) cos^B/ 1-sinB 1-sin^2B / 1-...2 Answers · Science & Mathematics · 08/07/2011

tan(x) +

**sec**(x) = 2cos(x) Change everything to sines and cosines... work, by plugging them into the original. tan(x) +**sec**(x) = 2cos(x) Test x = pi/6: tan(pi/6) +**sec**(pi/6) = 2cos(pi/6...6 Answers · Science & Mathematics · 24/07/2008

**sec**(b)csc(b) = 2csc(b)**sec**(b)csc(b) - 2csc(b) = 0 csc(b) * (**sec**(b) - 2) = 0 csc(b) = 0 1/sin(b...5pi/3 + 2pi * k b = (pi/3) * (1 + 6k) , (pi/3) * (5 + 6k) tan(t) +**sec**(t) = 1 sin(t)/cos(t) + 1/cos(t) = 1 (sin(t) + 1) / cos(t) = 1 sin(t...5 Answers · Science & Mathematics · 05/04/2012

∫

**sec**(x/2)dx Multiply numerator and denominator by (**sec**(x/2)+tan(x/2) and get ∫**sec**(x/2)dx= ∫{**sec**(x/2)(**sec**(x...x/2)tan(x/2)} dx= dt i.e dx =2dt/{**sec**²(x/2) +**sec**(x/2)tan(x/2)} Subsituting ∫{**sec**²(x...5 Answers · Science & Mathematics · 09/03/2011

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