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**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

∫

**sec**^4 x dx = ∫**sec**^2 x**sec**^2 x dx Integrate by parts dv=**sec**^2 x dx; v = tan x; u =**sec**^2 x ; du = 2**sec**^2 x tan x ∫ u dv = u v - ∫ v du...3 Answers · Science & Mathematics · 11/03/2014

Separate the integral to

**sec**(4x) tan (4x) since the derivative of**sec**x is**sec**x tan x, we'll need this...change tan^2 into**sec**^2 - 1 (trigonometry identity) Integral ((**sec**(4x) tan (4x)**sec**^2 (4x) (**sec**^2 (4x) - 1) dx) Use u substitute for**sec**(4x). Derivative...3 Answers · Science & Mathematics · 04/12/2011

To remember

**sec**-butyl, I always find it helpful to remember what the "**sec**" actually ...secondary carbon atoms (attached to two other carbons). The**sec**-butyl group is attached to the parent molecule via one of these "**sec**...6 Answers · Science & Mathematics · 07/07/2010

y =

**sec**(8^x) tan [ (1+e^x) /(1-e^x)] Apply the product rule: dy/dx = d/dx**sec**(8^x) tan [ (1+e^x) /(1-e^x)] +**sec**(8^x) d/dx... (3) into (2) d/dx tan [ (1+e^x)/(1-e^x)] =**sec**^2 [ (1+e^x)/(1-e^x)] times -2e^(2x) / (1...2 Answers · Science & Mathematics · 19/01/2012

**sec**(2 - 3x)^2 * sin(2 - 3x) / (1 -**sec**(2 - 3x))^(1/2) => (1/cos(2 - 3x)^2) * sin(2 - 3x) / (1 - 1/cos(2...1/3) * du / ((u + 1)^2 * sqrt(u)) u = tan(t)^2 du = 2 * tan(t) ***sec**(t)^2 * dt (-1/3) * 2 * tan(t) ***sec**(t)^2 * dt / ((1 + tan(t)^2)^2...1 Answers · Science & Mathematics · 19/02/2013

∫

**sec**(x√3) dx Let u = x√3 du = √3 dx 1/√3 ∫**sec**(u) du 1/√3 [ln|**sec**(u) + tan(u)| + C] ln|**sec**(x√3) + tan(x√3)| / √3 + C4 Answers · Science & Mathematics · 24/08/2011

∫

**sec**^3(x) dx = ∫**sec**(x)**sec**^2(x) dx from [ 0 to π/4 ] integrate by parts u =**sec**x du =**sec**(x) tan(x) dv =**sec**^2(x) dx v...π/4 ) tan(π/4) - 0 ] + (1/2) [ ln I**sec**(π/4) + tan(π/4) - ln I**sec**(0) + tan...2 Answers · Science & Mathematics · 03/01/2012

∫ ( 1/(1 -

**sec**(x)) dx ) We can't solve this directly; let's use identities. Multiply the top and bottom by (1 +**sec**(x)) gives us (1 +**sec**(x)) / (1 -**sec**²(x)) The bottom...3 Answers · Science & Mathematics · 18/03/2007

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