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∫

**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**^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

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

Hello, ∫ [

**sec**(3x) /tan(3x)] dx = let's rewrite it as: ∫**sec**(3x) [1/tan(3x)] dx...up with: (1/3) ln |csc(3x) - cot(3x)| + C in conclusion: ∫ [**sec**(3x) /tan(3x)] dx = (1/3) ln |csc(3x) - cot(3x)| + C I...1 Answers · Science & Mathematics · 05/10/2010

**sec**[arccot (-6)] the question is: what is the**sec**ant of that arc whose cot is (-6)? thus let [arccot (-6)] = y then cot y = - 6 now you have to rewrite**sec**y in terms of cot y, that is:**sec**²y = (1/cos²y...2 Answers · Science & Mathematics · 08/06/2008

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