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**sec**^8(x) * √(tan^3(x)) dx = ∫**sec**^6(x) ***sec**^2(x) * √(tan^3(x)) dx Note that 1 + tan^2(x...**sec**^2(x) * √(tan^3(x)) dx Now let z = tan(x) => dz =**sec**^2(x) dx = ∫(1 + z^2)^3 * z^(3/2) dz Now basically expand...2 Answers · Science & Mathematics · 31/03/2014

**sec**(2x) = [**sec**^2(x)] / [2 -**sec**^2(x)] Your first step would be to choose the more complex side. I'm... = 1 / cos(2x) Which, by definition is equal to RHS = 1 / cos(2x) =**sec**(2x) = LHS7 Answers · Science & Mathematics · 01/01/2007

y =

**sec**(xy^3) y' =**sec**(xy^3) tan(xy^3) d/dx (xy^3) y' =**sec**(xy^3) tan(xy^3) [ y^3 + x (3y^2) y' ] y' =**sec**(xy^3) tan(xy^3) [ y^3...1 Answers · Science & Mathematics · 01/10/2012

f(x) =

**sec**x = 1/cos x....................f(0) = 1 f'(x) = sin x/(cos² x...cos^5 x + 720 sin^6 x/cos^7 x f<6>(0) = 61 f(x) =**sec**x = 1 + x^2/2 + 5 x^4/4! + 61 x^6/6! +....**sec**x = 1 + x^2/2 + 5x^4/24...1 Answers · Science & Mathematics · 08/11/2008

...may not affect your reading of my answer. I dislike the

**SEC**because it seems that year in and year out, the**SEC**, more than any other...10 Answers · Sports · 20/08/2008

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**sec**x + tan x)cos x - sin x = 1 => In trigonometry,**sec**x = 1/cos x & tan x = sin x/cos x, then (1/cos x + sin x/cos x)cos x - sin x...3 Answers · Science & Mathematics · 20/10/2013

**sec**^2x sin^2x =**sec**^4x - (tan^4x +**sec**^2x) (1/cos^2x)sin^2x ==== (1/cos^4x) - (sin^4x/cos^4x) - (1/cos^2x...2 Answers · Science & Mathematics · 07/05/2019

∫ tan^2(x)

**sec**x dx ∫ [**sec**^2(x) - 1 ]**sec**x dx = ∫**sec**^3(x) dx - ∫**sec**x dx first one integrate by parts let u =**sec**x : du =**sec**x tan x dx dv =**sec**^2(x) dx : v = tan x ∫**sec**^3(x) dx =**sec**x...1 Answers · Science & Mathematics · 11/03/2009

Triple

**Sec**is used for Tequila Margarita, but you...on top of the Kahlua. Finally top with Triple**Sec**- also poured over the spoon. Serve :)2 Answers · Food & Drink · 12/03/2009

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**sec**² x - 1) /**sec**² x 1 - 1 /**sec**² x 1 - cos ² x (1 - cos x)(1 + cos x)3 Answers · Science & Mathematics · 09/04/2008

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